Powers and Roots: Identify the greater value

According to the given problem, whether it is discussed in addition or subtraction each of the terms that comes in its turn,

this is done within the framework of the order of operations, which states that multiplication precedes addition and subtraction, and that the preceding operations are performed before all others,

A. We start with the terms that are on the left in the given problem:

$\frac{1}{5}\cdot\big((5+3:3-8)^2:\sqrt{4}-2\big)$First, we simplify the terms that are in the denominators (the divisors) on which the multiplication operation is performed, this in accordance with the order of operations mentioned, we note that this term will change in terms that are in the denominators (the numerators) on which division is performed, therefore we start with simply the terms that are in these denominators, remembering that division precedes multiplication and subtraction, therefore the beginning will perform the division operation that is in this term and then perform the operations of multiplication and subtraction:

$\frac{1}{5}\cdot\big((5+3:3-8)^2:\sqrt{4}-2\big)\\ \frac{1}{5}\cdot\big((5+1-8)^2:\sqrt{4}-2\big)=\\ \frac{1}{5}\cdot\big((-2)^2:\sqrt{4}-2\big)\\$Since the results of the operations of multiplication and subtraction that are in the numerators the level will come out smoothly on these denominators, we continue and perform the strength on the term that is in these denominators, this within that we remember thatthe raising of any number (positive or negative) in a double strength will give a positive result, in contrast we will consider its numerical value of the other side that in strength he is the divisor that is in the term that within the denominators the divisors that were left (this within that we remember that in defining the root as strength, the root he is strength for everything):

$\frac{1}{5}\cdot\big((-2)^2:\sqrt{4}-2\big)=\\ \frac{1}{5}\cdot\big(4:2-2\big)\\$We continue and finish simply the term, we remember that division precedes subtraction and therefore the beginning will calculate the result of the division operation that in the term and then perform the operation of subtraction:

$\frac{1}{5}\cdot\big(2-2\big)=\\ \frac{1}{5}\cdot0=\\ 0$In the last stage we performed the doubling that was left (it is the doubling that is performed on the term that in the denominators), this within that we remember thatthe result of doubling any number (different from zero) in zero here zero,

$$We finished simply the term that is on the left in the given problem, we will summarize the stages of simply:

We received that:

$\frac{1}{5}\cdot\big((5+3:3-8)^2:\sqrt{4}-2\big)\\ \frac{1}{5}\cdot\big((-2)^2:\sqrt{4}-2\big)\\ \frac{1}{5}\cdot\big(2-2\big)=\\ 0$

B. We continue and simplify the term that is on the right in the given problem:

$8-(3^2+1)\cdot\frac{1}{10}$In this part to do in the first part simplify the term within the framework of the order of operations,

In this term a doubling that is performed on the term that in the denominators, therefore we simplify first this term, this is done in accordance with the order of operations mentioned, therefore we start from considering its numerical value that in strength that in this term and then perform the operation of multiplication:

$8-(3^2+1)\cdot\frac{1}{10} =\\ 8-(9+1)\cdot\frac{1}{10}=\\ 8-10\cdot\frac{1}{10}\\$We continue and simplify the term that was received in the first stage, we remember in this that multiplication and division precede addition and subtraction, therefore the beginning will perform the doubling in the break, this within that we remember that the doubling in the break means the doubling in the amount of the break, then perform the operation of division that of the break, this is done by appointment, in the last stage will perform the operation of subtraction that remained:

$8-10\cdot\frac{1}{10}=\\ 8-\frac{10\cdot1}{10}=\\ 8-\frac{\not{10}}{\not{10}}=\\ 8-1=\\ 7$We finished simply the term that is on the right in the given problem, we will summarize the stages of simply:

We received that:

$8-(3^2+1)\cdot\frac{1}{10} =\\ 8-10\cdot\frac{1}{10}\\ 7$We return now to the original problem, and we will present the results of simply the terms that were reported in A and in B:

$\frac{1}{5}\cdot\big((5+3:3-8)^2:\sqrt{4}-2\big)\text{ }_{\textcolor{red}{——}\text{\textcolor{red}{ }}}8-(3^2+1)\cdot\frac{1}{10} \\ \downarrow\\ 0\text{ }\text{\textcolor{red}{\_\_}}\text{ }7$Therefore the correct answer here is answer A.

According to the given problem, whether it's an addition or subtraction each of the digits that come up separately,

this is done within the framework of the order of operations, which states that multiplication precedes addition and subtraction, and that the preceding operations are done before division and subtraction, and that the preceding operations are done before all,

A. We start with the digits that are left in the given problem:

$-3+(\sqrt{100}-3^2-1^{14}):30+3$First, we simplify the digits that are in the denominators on which the division operation takes place, this is done in accordance with the order of operations mentioned, keeping in mind that multiplication precedes subtraction, therefore, we first calculate their numerical values of the numerators in the multiplication (this within the definition of the root as a power, the root of which is a power for everything), subsequently we perform the subtraction operation which is within the denominators and finally we perform the division operation that takes place on the denominators:

$-3+(\sqrt{100}-3^2-1^{14}):30+3 =\\ -3+(10-9-1):30+3 =\\ -3+0:30+3 =\\ -3+0+3 \\$In the last step we mentioned that dividing the number 0 by any number (different from zero) will yield the result 0, we continue with the simple digits we received in the last step and perform the addition operation:

$-3+0+3 =\\ 0$We finish the simple digits that are left in the given problem, we summarize the steps of the simplification:

We received that:

$-3+(\sqrt{100}-3^2-1^{14}):30+3 =\\ -3+0:30+3 =\\ 0$

B. We continue and simplify the digits that are right in the given problem:

$6^2:6\cdot(3-2)-6$In this part, in the first step we simplify the digits within the framework of the order of operations,

In this digit, multiplication takes place on digits in the denominators, therefore, we first simplify this digit, in the process we calculate the numerical values of the numerator in the multiplication which is the divisor in the first digit from the left in the given digit:

$6^2:6\cdot(3-2)-6 =\\ 36:6\cdot1-6 \\$We continue and remember that multiplication and division precede addition and subtraction, keeping in mind that there is no predefined precedence between multiplication and division operations originating from the order of operations mentioned, therefore, we calculate the numerical values of the numerator in the first digit from the left (including the multiplication and division operations) within the execution of one operation after another according to the order from left to right (this is the natural order of operations), subsequently, we complete the calculation within the execution of the subtraction operation:

$36:6\cdot1-6 =\\ 6\cdot1-6 =\\ 6-6 =\\ 0$We finish the simple digits that are right in the given problem, we summarize the steps of the simplification:

We received that:

$6^2:6\cdot(3-2)-6 =\\ 36:6\cdot1-6 =\\ 0$We return to the original problem, and we present the results of the simplifications reported in A and B:

$-3+(\sqrt{100}-3^2-1^{14}):30+3\text{ }\text{\textcolor{red}{\_\_}}\text{ }6^2:6\cdot(3-2)-6 \\ \downarrow\\ 0\text{ }\text{\textcolor{red}{\_\_}}\text{ }0$As a result, we have that:

$0 \text{ }\textcolor{red}{=}0$Therefore, the correct answer here is answer B.

Simplify each of the following expressions separately :

a. Let's start with the expression on the left:

$(9^2-2\sqrt{81})+4^3:2^3$

Note that in this expression the parentheses are meaningless given that no mathematical operation is applied to them, therefore we can remove them, whilst writing the division operation as a fraction:

$(9^2-2\sqrt{81})+4^3:2^3=\\ 9^2-2\sqrt{81}+\frac{4^3}{2^3}$

Note that the number 4 is a power of the number 2:

$4=2^2$

Therefore, we can have all terms with identical bases in the fraction and use the power rule for dividing terms with identical bases:

$\frac{a^m}{a^n}=a^{m-n}$

Remember the power rule for power of a power:

$(a^m)^n=a^{m\cdot n}$

Apply these two rules in handling the fraction in the expression mentioned above (which is the third term from the left):

$\frac{4^3}{2^3} =\frac{(2^2)^3}{2^3} =\frac{2^{2\cdot3}}{2^3} =\frac{2^6}{2^3}=2^{6-3} =2^3=8$

Where in the first stage we replaced the number 4 with a power of 2, in the next stage we applied the power rule for power of a power mentioned earlier, and then we simplified the expression in the fraction's numerator and applied the power rule for dividing terms with identical bases. We completed the calculation by simplifying the resulting expression,

Let's return to the expression and complete its calculation by substituting what we've obtained so far:

$9^2-2\sqrt{81}+\frac{4^3}{2^3} \\ \downarrow\\ 9^2-2\sqrt{81}+8=\\ 81-2\cdot9+8=\\ 81-18+8=71$

b. Let's continue with the expression on the right:

$(8^2+2\sqrt{64})-3^2$

Here too we notice that the parentheses are meaningless since no mathematical operation is applied to them, therefore, similar to the previous part, we'll remove them:

$8^2+2\sqrt{64}-3^2$

notice that in this expression there's no special use of power rules therefore we'll continue with direct simplification through numerical calculation only:

$8^2+2\sqrt{64}-3^2 =\\ 64+2\cdot8-9=\\ 64+16-9=71$

Let's return to the original problem and substitute the results of both solution parts detailed so far in a and b in place of the expressions on the right and left:

$(9^2-2\sqrt{81})+4^3:2^3\text{ }\textcolor{red}{_{——\text{ }}}(8^2+2\sqrt{64})-3^2 \\ \downarrow\\ 71\text{ }\textcolor{red}{_{——\text{ }}}71$

Therefore it's clear that equality exists, meaning that:

$71\text{ }\textcolor{red}{{=\text{ }}}71$

The correct answer is answer c.

To solve the given problem and determine whether it is an equality or inequality, we need to simplify the expressions on either side.

We can deal with each of them separately and simplify them, however, a more efficient way of working will be to deal with the more complex parts of these expressions separately, that is, with the expressions containing roots,

It's important to emphasize that in generalwe want to learn to solve without using a calculator by using our algebraic tools and the laws of exponents. Let's begin:

a. Let's start with the first part of the expression on the right:

$\sqrt{10}+\sqrt{2}\cdot\sqrt{5}$

(We'll focus on the expression inside the parentheses first and then continue outwards).

Let's recall two laws of exponents:

a.1: Defining a root as an exponent:

$\sqrt[n]{a}=a^{\frac{1}{n}}$

a.2: The law for an exponent applied to parentheses containing multiplication, but in the opposite direction:

$x^n\cdot y^n= (x\cdot y)^n$

Usually we would replace roots with exponents, but for now we won't do that. In the meantime just understand that according to the law of defining a root as an exponent mentioned in a.1, the root is actually an exponent and therefore all laws of exponents apply to it, especially the law of exponents mentioned ina.2. Let's apply this understanding to the expression in question:

$\sqrt{10}+\sqrt{2}\cdot\sqrt{5} = \sqrt{10}+\sqrt{2\cdot5}= \sqrt{10}+\sqrt{10}$

In the first step, we notice that the second term (i.e., the product of roots) is actually a product between two terms raised to the same exponent (which is half the power of the square root). Therefore using to the law of exponents mentioned in a.2, we can multiply the bases of the terms under the same square root, and in the next stage we simplify the expression in the root.

From here we note that this expression can be factored by a common factor:

$\sqrt{10}+\sqrt{10}=(1+1)\sqrt{10}=2\cdot\sqrt{10}$

We used the commutative property of multiplication, that is, the common factor we took out:$\sqrt{10}$We took out to the right of the parentheses (instead of to their left),

Let's continue to the second problematic expression on the right:

$\sqrt{48}\cdot\sqrt{3}$

And we'll deal with it the same way as in the previous part:

$\sqrt{48}\cdot\sqrt{3} = \sqrt{48\cdot3}=\sqrt{144}$

Again, in the first step, we notice that the expression (i.e., the product of roots) is actually a product between two terms raised to the same exponent (which is half the power of the square root), therefore according to the law of exponents mentioned in a.2, we can multiply the bases of the terms under the same exponent, and in the next step we simplify the expression in the root,

From here we'll return to the original expression in the problem (i.e., the expression on the right) and calculate it in full, using the simplified expressions we got above:

$4^2(\sqrt{10}+\sqrt{2}\cdot\sqrt{5})^2+\sqrt{48}\cdot\sqrt{3} = \\ 4^2(2\cdot\sqrt{10})^2+\sqrt{144}$

We just substitute what we calculated earlier in the two parts above - in place of the expression in parentheses and in place of the second term.

INext, let's recall again the law of exponents mentioned above in a.2, that is, the law for an exponent applied to parentheses containing a product but in the normal way.

Let's apply this law to the expression we got in the last stage:

$4^2(2\cdot\sqrt{10})^2+\sqrt{144} =4^2\cdot2^2\cdot(\sqrt{10})^2+\sqrt{144}=\\ 16\cdot4\cdot10+12$

We apply the law of exponents mentioned in a.2 above and applied the exponent to each of the multiplication terms in the parentheses.

Next, we apply the exponent to the square root while remembering that these are actually two inverse operations and therefore cancel each other out and simultaneously simplify the other terms by applying the square root to the second term on the left, and the exponents to the first term.

Let's finish solving. We got that the expression is:

$16\cdot4\cdot10+12 =640+12=652$

Let's summarize this part:

We got that the expression on the right is:

$4^2(\sqrt{10}+\sqrt{2}\cdot\sqrt{5})^2+\sqrt{48}\cdot\sqrt{3}=\\ 4^2(2\cdot\sqrt{10})^2+\sqrt{144} =16\cdot4\cdot10+12 =\\ 652$

b. Let's continue to the expression on the left and start by writing it in the standard fraction notation while keeping in mind the order of operations, that is - parentheses, exponents, multiplication/ division, addition/ subtraction.

$3^3-5^2:(2^2+1)= 3^3-\frac{5^2}{2^2+1}$

Since the division here refers to the entire expression in parentheses, we inserted it in its entirety into the denominator of the fraction,

Let's simplify this expression. We'll focus on the fraction, but first let's recall the law of multiplying exponents with identical bases:

b.1:

$\frac{a^m}{a^n}=a^{m-n}$Let's simplify the expression. We'll start by simplifying the numerator of the fraction and continue by applying the law of exponents mentioned above:

$3^3-\frac{5^2}{2^2+1} =3^3-\frac{5^2}{4+1} =3^3-\frac{5^2}{5} =\\ 3^3-5^{2-1}=3^3-5$

In the first and second parts, we simplified the numerator of the fraction, in the third part we applied the law of exponents mentioned in b.1 and in the following parts we simplified the resulting expression.

$$Let's finish the calculation:

$3^3-5 =27-5=22$

And to summarize:

$3^3-5^2:(2^2+1)= 3^3-\frac{5^2}{2^2+1} =\\ 3^3-\frac{5^2}{5} = 3^3-5^{2-1}=3^3-5=22$

Now let's return to the original problem and replace what we got for a' and b':

$3^3-5^2:(2^2+1)_{\textcolor{red}{_{——}}\text{ }}4_{\text{\textcolor{red}{ }}}^2(\sqrt{10}+\sqrt{2}\cdot\sqrt{5})^2+\sqrt{48}\cdot\sqrt{3} \\ \downarrow\\ 22\text{ }\text{\textcolor{red}{\_\_}}\text{ }652$

Therefore it's clear that this is not an equality but an inequality and that the expression on the left is smaller than the expression on the right, meaning that:

$22\text{ }\text{\textcolor{red}{<}}\text{ }652$

Therefore the correct answer is answer a'.

To solve the given problem and determine whether it is an equation or inequality, we need to simplify each of the algebraic expressions.

We can deal with each of them separately and simplify them. However, in this case, a more efficient way is be to deal with the more complex parts of these expressions separately, that is - the expressions in parentheses - the expressions with roots.

It's important to emphasize that in general we want to solve without a calculator, using only our algebraic tools and laws of exponents. Let's begin:

a. We'll start with the problematic part in the left expression:

$\sqrt{6}+\sqrt{2}\cdot\sqrt{3}$(We'll focus on the expression inside the parentheses first and then move outwards),

Let's recall two laws of exponents:

a.1: Defining a root as an exponent:

$\sqrt[n]{a}=a^{\frac{1}{n}}$

a.2: The law of applying exponents to parentheses containing a product, but in the opposite direction:

$x^n\cdot y^n= (x\cdot y)^n$

Usually we replace roots with exponents, but for now we won't do that. We'll just understand that according to the law of defining roots as an exponent mentioned in a.1, the root is actually an exponent and therefore all laws of exponents apply to it, especially the law of exponents mentioned in a.2 , So we'll apply this understanding to the expression in question:

$\sqrt{6}+\sqrt{2}\cdot\sqrt{3} = \sqrt{6}+\sqrt{2\cdot3}= \sqrt{6}+\sqrt{6}$

In the first stage, we notice that the second term (i.e., the product of roots) is actually a product between two terms raised to the same exponent (which is half the power of the square root). Therefore, according to the law of exponents mentioned in a.2 , we can combine the bases of the terms as a product with the same exponent , and in the next stage we simplify the expression under the root,

From here we notice that we can simplify this expression by a using common factor:

$\sqrt{6}+\sqrt{6}=(1+1)\sqrt{6}=2\cdot\sqrt{6}$

We used the commutative property of multiplication to move the common factor we took out ($\sqrt{6}$ ) we put it on the right of the parentheses (instead of to their left) so our expression with be clearer.

Now, we'll return to the original expression in the problem (i.e., the expression on the left) and calculate in full, using the simplification from above:

$9\cdot(\sqrt{6}+\sqrt{2}\cdot\sqrt{3})^2+2^2:2= 9\cdot(2\cdot\sqrt{6})^2+\frac{2^2}{2^1}$

We substitute what we calculated above in place of the expression in parentheses and write the division operation for the last term on the left as a fraction.

In the next stage let's recall another law of exponents:

a.3: The law of dividing exponents with equal bases:

$\frac{a^m}{a^n}=a^{m-n}$And let's recall again the law of exponents mentioned above in a.2, that is, the law of an exponent applied to parentheses containing a product, but in the normal direction.

Let's apply these two laws of exponents to the expression we got in the last stage:

$9\cdot(2\cdot\sqrt{6})^2+\frac{2^2}{2^1} =9\cdot2^2\cdot(\sqrt{6})^2+2^{2-1}=9\cdot4\cdot6+2^1$

In the first stage we apply the law of exponents mentioned in a.2 above and apply the exponent to each of the multiplied terms in the parentheses.

Then, we apply the law of exponents mentioned in a.3 to the second term on the left. To make things clearer, we put the root in parentheses, but this just for convenience.

In the next stage we square the square root while remembering that these are actually two inverse operations and therefore cancel each other out and we simplify the rest of the terms.

Let's finish the calculation. We got that the expression is:

$9\cdot4\cdot6+2^1 =216+2=218$

Let's summarize:

We got that the left expression is:

$9\cdot(\sqrt{6}+\sqrt{2}\cdot\sqrt{3})^2+2^2:2= 9\cdot(2\cdot\sqrt{6})^2+\frac{2^2}{2^1} =\\ 9\cdot4\cdot6+2^1 =218$

b. Let's continue to the expression on the right and as before,we'll start with the problematic part, that is, the expression inside the parentheses with the roots:

$\sqrt{2}+2\sqrt{2}$Just as in the previous part, we can factor this expression:

$\sqrt{2}+2\sqrt{2} =(1+2)\sqrt{2}=3\cdot\sqrt{2}$

Again we use the commutative property of multiplication and the common factor-$\sqrt{2}$ we choose to take out outside the parentheses - to their right.

Next, we simplify the expression in parentheses.

We'll return to the full expression on the right and substitutewhat we got :

$5^2\cdot(\sqrt{2}+2\sqrt{2})^2 =5^2\cdot(3\cdot\sqrt{2})^2$

We'll continue and simplify this expression. Again, we wil apply the law of exponents mentioned earlier in a.2 (in its normal direction) and we keep in mind that the square root and the square exponent are inverse operations and therefore cancel each other out:

$5^2\cdot(3\cdot\sqrt{2})^2 =5^2\cdot3^2 \cdot(\sqrt{2})^2=25\cdot9\cdot2$

First, according to the law of exponents mentioned in a.2 we apply the exponent to each of the multiplied terms in parentheses, and then we simplify the resulting expression while we keep in mind that the square root and square exponent are inverse operations.

Let's finish the calculation:

$25\cdot9\cdot2=450$And to summarize , we got that:

$5^2\cdot(\sqrt{2}+2\sqrt{2})^2 =5^2\cdot(3\cdot\sqrt{2})^2 =\\ 5^2\cdot3^2 \cdot(\sqrt{2})^2=25\cdot9\cdot2 =450$

Now let's return to the original problem and substitute what we got in a and b:

$9\cdot(\sqrt{6}+\sqrt{2}\cdot\sqrt{3})^2+2^2:2\text{ }\text{\textcolor{red}{\_\_}}\text{ }5^2\cdot(\sqrt{2}+2\sqrt{2})^2 \\ \downarrow\\ 218\text{ }\text{\textcolor{red}{\_\_}}\text{ }450$Therefore it's clear that this is not an equality but an inequality and that the expression on the left is smaller than the expression on the right ,that is:

$218\text{ }\text{\textcolor{red}{<}}\text{ }450$Therefore the correct answer is answer a.

To solve the problem, we need to carefully evaluate both expressions on either side of the _ (underscored) red dash, $\textcolor{red}{—}$. Let's go through the calculations step-by-step, using the order of operations.

Step 1: Evaluate each component in the expressions separately:

• $10^2$: 10 squared is 100.

• $\sqrt{16}$: The square root of 16 is 4.

• $2^2$: 2 squared is 4.

• $\sqrt{49}$: The square root of 49 is 7.

• $3^2$: 3 squared is 9.

• $2^3$: 2 cubed is 8.

Step 2: Substitute the evaluated components back into the expression and reduce:

• Left Expression: $(100 : 4 - 4 \cdot 6)^{100}$

  Calculate inside the parentheses using division first: $100 : 4 = 25$

  Then calculate the multiplication: $4 \cdot 6 = 24$

  Subtract the results: $25 - 24 = 1$

  Raise to the 100th power: $1^{100} = 1$

• Right Expression: $7 : 7 + 9 - 8$

  Calculate the division: $7 : 7 = 1$

  Then add: $1 + 9 = 10$

  Finally, subtract: $10 - 8 = 2$

After evaluating, we're left with comparing $1$ on the left side and $2$ on the right side.

Conclusion: $1 \lt 2$. Therefore, the appropriate sign to mark is $\lt$.

Let's deal with each of the expressions on the right and left separately:

a. We'll start with the expression on the left:

$\frac{\sqrt{16}\cdot(6^2-7\cdot3)}{(4^3-2\cdot3^3):2}$Since it's very messy, we'll start by organizing it.

First, we'll isolate the expression in the denominator of the fraction (which is currently shown as a division operation), and treat it as a fraction by itself:

$\frac{\sqrt{16}\cdot(6^2-7\cdot3)}{(4^3-2\cdot3^3):2} =\\ \frac{\sqrt{16}\cdot(6^2-7\cdot3)}{\big(\frac{4^3-2\cdot3^3}{2}\big)}$Note that the division operation in the denominator applies to the entire expression in parentheses (meaning that the entire expression in parentheses in the denominator is divided by 2), therefore the entire expression that was in parentheses becomes the numerator of the new fraction.

Additionally, we put the fraction in the denominator in parentheses, and this is to be able to distinguish between the main fraction line (of the large, main fraction) and that of the secondary fraction line (of the fraction in the denominator),

Remember that division is multiplication by the reciprocal number, and also that we get the reciprocal of a fraction by swapping the numerator and denominator, that is - mathematically, we'll perform:

$\frac{a}{\big(\frac{x}{y}\big)}=\frac{\big(\frac{a}{1}\big)}{\big(\frac{x}{y}\big)}=\frac{a}{1}\cdot\frac{y}{x}$In the first part we remembered that any number can be represented as that same number divided by 1, and then we converted the division operation in the fraction to a multiplication operation by the reciprocal fraction. we'll apply this to the expression we received in the last stage:

$\frac{a}{\textcolor{blue}{\big(\frac{x}{y}\big)}}=\frac{\big(\frac{a}{1}\big)}{\textcolor{blue}{\big(\frac{x}{y}\big)}}=\frac{a}{1}\cdot\textcolor{blue}{\frac{y}{x}} \\ \downarrow\\ \frac{\sqrt{16}\cdot(6^2-7\cdot3)}{\textcolor{blue}{\big(\frac{4^3-2\cdot3^3}{2}\big)}} = \frac{\big(\frac{\sqrt{16}\cdot(6^2-7\cdot3)}{1}\big)}{\textcolor{blue}{\big(\frac{4^3-2\cdot3^3}{2}\big)}}=\frac{\sqrt{16}\cdot(6^2-7\cdot3)}{1}\cdot\textcolor{blue}{\frac{2}{4^3-2\cdot3^3} }$From here we'll continue as usual and perform the multiplication operation between the fractions, remembering that when multiplying fractions we multiply numerator by numerator and denominator by denominator (and keep the original fraction line):

$\frac{\sqrt{16}\cdot(6^2-7\cdot3)}{1}\cdot\frac{2}{4^3-2\cdot3^3} =\frac{\sqrt{16}\cdot(6^2-7\cdot3)\cdot2}{1\cdot(4^3-2\cdot3^3)}$We have thus finished organizing the expression.

We did carefully while paying attention to the separation between the main and secondary fraction lines (and so on),

We'll continue and use the distributive property of multiplication, and the fact that multiplying any number by one will give us the number itself (and we'll get rid of the parentheses in the fraction's denominator):

$\frac{\sqrt{16}\cdot(6^2-7\cdot3)\cdot2}{1\cdot(4^3-2\cdot3^3)} =\frac{\sqrt{16}\cdot2\cdot(6^2-7\cdot3)}{4^3-2\cdot3^3}$Here we rearranged the expression in the fraction's numerator using the distributive property mentioned above while keeping the parentheses and treating them together with the expression inside them as one unit,

We'll continue and calculate the value of the fraction we received in the last step by finding the values of the expressions, and while being careful about the order of operations.

We'll start by calculating the expression in parentheses in the numerator:

$\frac{\sqrt{16}\cdot2\cdot(6^2-7\cdot3)}{4^3-2\cdot3^3} =\frac{\sqrt{16}\cdot2\cdot(36-21)}{64-2\cdot27}= \frac{\sqrt{16}\cdot2\cdot15}{64-54} =\frac{\sqrt{16}\cdot2\cdot15}{10}$While simplifying the expression in parentheses in the fraction's numerator we simplified the expression in the fraction's denominator.

We'll continue and finish reducing this expression by finding the value of the root in the fraction's numerator, and then performing the division operation of the fraction itself:

$\frac{\sqrt{16}\cdot2\cdot15}{10} =\frac{4\cdot2\cdot15}{10} =\frac{120}{10}=12$

Let's summarize:

$\frac{\sqrt{16}\cdot(6^2-7\cdot3)}{(4^3-2\cdot3^3):2} = \frac{\sqrt{16}\cdot(6^2-7\cdot3)}{\textcolor{blue}{\big(\frac{4^3-2\cdot3^3}{2}\big)}} =\\ \frac{\sqrt{16}\cdot(6^2-7\cdot3)}{1}\cdot\textcolor{blue}{\frac{2}{4^3-2\cdot3^3} } = \frac{\sqrt{16}\cdot2\cdot(6^2-7\cdot3)}{4^3-2\cdot3^3} =12$

b. We'll continue to the expression on the right:

$\frac{\frac{9}{\sqrt{81}}-(1-\frac{2}{3})^2}{2^4:4^2}$Similar to the previous expression, this expression is also messy, therefore, we'll organize it first. Unlike before, here we'll start by simplifying the expression in parentheses in the fraction's numerator, that is, simplifying the expression:

$1-\frac{2}{3}$We'll perform this subtraction first by converting the 1 into a fraction and then by finding a common denominator.

First we'll represent the whole number as that same number divided by 1 (which is always possible and advisable to do):

$1-\frac{2}{3} =\frac{1}{1}-\frac{2}{3}$From here we can see that the common denominator is the number 3. So we'll convert the fractions to fractions with common denominator and simplify the expression.

Remember that when subtracting fractions with the same denominator we subtract the numerators and keep the denominator:

$\frac{1}{1}-\frac{2}{3} =\frac{1\cdot3-2\cdot1}{3}=\frac{3-2}{3}=\frac{1}{3}$In the first stage we put the two fractions on a fraction line with a common denominator as described above, and in the next stage we simplified the expression that we got.

We'll now return to the original expression and substitute the result we got for the expression in parentheses:

$\frac{\frac{9}{\sqrt{81}}-(1-\frac{2}{3})^2}{2^4:4^2} = \frac{\frac{9}{\sqrt{81}}-(\frac{1}{3})^2}{2^4:4^2}$We'll continue and simplify the expression in the fraction's numerator (the main one) while keeping in mind the law of exponents with the same base:

$\big(\frac{a}{c}\big)^n=\frac{a^n}{c^n}$We'll apply this law of exponents to the expression in the main fraction's numerator, as we simplify the second fraction in the main fraction's numerator by calculating the value of the root and simplifying the expression:

$\frac{\frac{9}{\sqrt{81}}-(\frac{1}{3})^2}{2^4:4^2} = \frac{\frac{9}{9}-\frac{1^2}{3^2}}{2^4:4^2} =\\ \frac{1-\frac{1}{9}}{2^4:4^2}$

In the first stage we applied the above-mentioned law of exponents which states that for exponents with the same base applied to parentheses containing a sum of terms, we apply the exponent separately to both the numerator and denominator (of that same fraction).

We applied the same law to the second term in the main fraction's numerator, while we simplified the first expression in the main fraction's numerator by finding the value of the root and simplifying the expression (and we remembered that dividing any number by itself will always give the result 1),

We'll continue by dealing with the expression in the main fraction's numerator:

We have another subtraction operation between a whole number and a fraction, we'll perform it separately:

$1-\frac{1}{9}=\frac{1}{1}-\frac{1}{9}=\frac{1\cdot9-1\cdot1}{9}=\frac{9-1}{9}=\frac{8}{9}$We repeat what was described earlier: we perform the subtraction operation between the two fractions after finding the common denominator, the number 9, and simplifying the expression in the numerator,

We'll now return to the original expression and substitute the result.

Let's summarize what we've done so far:

$\frac{\frac{9}{\sqrt{81}}-(1-\frac{2}{3})^2}{2^4:4^2} = \frac{\frac{9}{\sqrt{81}}-(\frac{1}{3})^2}{2^4:4^2} =\\ \frac{1-\frac{1}{9}}{2^4:4^2} =\frac{\big(\frac{8}{9}\big)}{2^4:4^2}$

Again, we used parentheses, this time in the fraction's numerator to emphasize the main fraction line in the expression.

We'll continue and write the division operation in the fraction's denominator (the main one) as a fraction:

$\frac{\big(\frac{8}{9}\big)}{2^4:4^2} =\frac{\big(\frac{8}{9}\big)}{\big(\frac{2^4}{4^2}\big)}$We used parentheses again, this time in the fraction's denominator to emphasize the main fraction line in the expression.

Next we'll apply again the laws of exponents - first we'll replace the number 4 with an exponent of the number 2:

$4=2^2$This is to get expressions with the same base in the fraction which is in the main fraction's denominator, we'll now deal with the expression which is in the main fraction's denominator separately:

$\frac{2^4}{4^2}=\frac{2^4}{(2^2)^2}$

We'll deal with this expression and remember the law of exponents:

$(a^m)^n=a^{m\cdot n}$We'll apply this law to the denominator of the fraction we are dealing with:

$\frac{2^4}{(2^2)^2} =\frac{2^4}{2^{2\cdot2}}=\frac{2^4}{2^{4}} =1$In the first part we apply the law of exponents and in the following steps we simplify the expression we got.

In the last step it was possible to use the law of exponents for terms with the same base and get the same result, but here it's simpler to divide.

We'll return to the original expression and summarize what we got, while we substitute the result of the last calculation and simplify:

$\frac{\frac{9}{\sqrt{81}}-(1-\frac{2}{3})^2}{2^4:4^2} = \frac{\frac{9}{\sqrt{81}}-(\frac{1}{3})^2}{2^4:4^2} =\\ \frac{1-\frac{1}{9}}{2^4:4^2} =\frac{\big(\frac{8}{9}\big)}{2^4:4^2}= \frac{\big(\frac{8}{9}\big)}{\big(\frac{2^4}{4^2}\big)} =\\ \frac{\big(\frac{8}{9}\big)}{\big(\frac{2^4}{(2^2)^2}\big)} = \frac{\big(\frac{8}{9}\big)}{\big(\frac{2^4}{2^{4}} \big)}=\frac{\big(\frac{8}{9}\big)}{1 }=\frac{8}{9}$Keep in mind that dividing any number by one will give us that same number.

We have now finished dealing with the expression on the right, we'll return to the original problem and substitute the results of the expressions on the left and right which were calculated in a' and b' respectively:

$\frac{\sqrt{16}\cdot(6^2-7\cdot3)}{(4^3-2\cdot3^3):2}\text{ }_{\textcolor{red}{—}\text{ }}\frac{\frac{9}{\sqrt{81}}-(1-\frac{2}{3})^2}{2^4:4^2} \\ \downarrow\\ 12\text{ }_{\textcolor{red}{—}\text{ }}\frac{8}{9}$To determine which expression is larger we can present the expression on the left as a fraction with denominator 9, but since here the expression on the left is clearly larger than the number 1, while the expression on the right is smaller than the number 1, we can conclude that:$12>1>\frac{8}{9}$And therefore it is certainly true that:

$12\text{ }{\textcolor{red}{>}\text{ }}\frac{8}{9}$That is, the correct answer is answer b.

Let's deal with each of the expressions, the one on the left and the one on the right separately:

A. We'll start with the expression on the left:

$\frac{3^2\cdot(8-2\cdot3)^3}{(5^2\cdot3-72)\cdot\sqrt{16}}$We'll simplify the expressions in the numerator and denominator while being careful to follow the order of operations.

Remember that we solve parentheses first, and then exponents, then multiplication and division, and lastly addition and subtraction (when dealing with a fraction, the operation is actually the numerator of the fraction (in parentheses) divided by the denominator of the fraction (in parentheses), so we deal with them separately while maintaining the main fraction line:

If so, we'll start by dealing with the numerator of the fraction, where we'll first simplify what's in the parentheses and then raise that to the power that's outside the parentheses, and finally multiply them:

$\frac{3^2\cdot(8-2\cdot3)^3}{(5^2\cdot3-72)\cdot\sqrt{16}} = \frac{3^2\cdot(8-6)^3}{(5^2\cdot3-72)\cdot\sqrt{16}}=\\ \frac{3^2\cdot2^3}{(5^2\cdot3-72)\cdot\sqrt{16}}= \frac{9\cdot8}{(5^2\cdot3-72)\cdot\sqrt{16}}$

We'll continue in the same way- simplifying the expression in the denominator of the fraction.

Again, we'll start by simplifying the expression in parentheses and then finding the root:

$\frac{9\cdot8}{(5^2\cdot3-72)\cdot\sqrt{16}} =\frac{9\cdot8}{(25\cdot3-72)\cdot\sqrt{16}} =\\ \frac{9\cdot8}{(75-72)\cdot\sqrt{16}} =\frac{9\cdot8}{3\cdot4}$

Now we can do the multiplications in the numerator and denominator and divide between the results. Notice that we can reduce the fraction and this is because between all the terms both in the numerator and its denominator there is a common factor and because the terms in the numerator are factors of the terms in the denominator, in other words - the number 8 can be divided by 4 and the number 9 can be divided by 3:

$\frac{\not{9}\cdot\not{8}}{\not{3}\cdot\not{4}}=3\cdot2=6$

Let's summarize what we've done up until now:

$$$\frac{3^2\cdot(8-2\cdot3)^3}{(5^2\cdot3-72)\cdot\sqrt{16}} = \frac{9\cdot8}{(5^2\cdot3-72)\cdot\sqrt{16}} =\frac{9\cdot8}{3\cdot4} =6$And we've finished dealing with the left term.

B. Let's continue and deal with the expression on the right:

$\frac{\frac{3}{2^3}-\big(1-\frac{1}{2}\big)^3}{1^4-0.5}$Here we'll notice that in the numerator there is a subtraction operation inside parentheses that are raised to a power, and that the whole expression in the numerator is actually a subtraction between fractions.

Therefore let's we'll deal with the numerator of the fraction separately, meaning we'll simplify the expression:

$\frac{3}{2^3}-(1-\frac{1}{2})^3$We'll start by calculating the subtraction in parentheses:

$1-\frac{1}{2}=\frac{1}{1}-\frac{1}{2}=\frac{1\cdot2-1\cdot1}{2}=\\ \frac{2-1}{2}=\frac{1}{2}$

We performed the subtraction by finding a common denominator (2) and converting each of the fractions by asking "By how much did we multiply the current denominator to get the common denominator?" Then we simplified the expression that was obtained in the denominator of the fraction.

Let's continue simplifying the main numerator of the fraction, we'll substitute the result we got in the last step in the expression we're dealing with:

$\frac{3}{2^3}-\big(1-\frac{1}{2}\big)^3 = \frac{3}{2^3}-\big(\frac{1}{2}\big)^3$Now let's remember the law of exponents for raising parentheses containing a sum of terms to a power:

$\big(\frac{a}{c}\big)^n=\frac{a^n}{c^n}$We'll apply this law of exponents to the expression we got in the last stage, at the same time we'll find the value of the expression in the denominator of the fraction of the first term from the left:

$\frac{3}{2^3}-\big(\frac{1}{2}\big)^3 =\frac{3}{8}-\frac{1^3}{2^3}=\frac{3}{8}-\frac{1}{8} =\\ \frac{3-1}{8}=\frac{2}{8}$After applying the exponents law we subtracted the fractions. Here, since both fractions that were involved already had the same denominator (it's the common denominator here) we could just simplify the expression in the numerator of the fraction.

Let's summarize then what we've done so far in dealing with the right term, we got that:

$\frac{\not{2}}{\not{8}} =\frac{1}{4}$

Let's continue with the expression in the denominator of the fraction on the right:

$\frac{3}{2^3}-\big(1-\frac{1}{2}\big)^3 = \frac{3}{2^3}-\big(\frac{1}{2}\big)^3 =\\ \frac{3}{8}-\frac{1}{8} =\frac{3-1}{8}=\frac{2}{8} = \frac{1}{4}$It's preferable here to work with simple fractions (or mixed numbers) and not decimal fractions, so we'll represent the second term from the left as a simple fraction, while keepig in mind the definition of a decimal fraction.$1^4-0.5$In the first step we used our uderstandig of decimal fractions to convert five tenths to their fractional form, then we reduced the fraction that we got.

Let's return to the expression in question and substitute the fractional form we calculated above, let's remember that raising the number 1 to any power will always give the result 1:

$0.5=\frac{\not{5}}{\not{10}}=\frac{1}{2}$Here we need to do a subtraction between a whole number and a fraction, a calculation identical to the calculation already done in this solution in the previous stage (in B).

Let's summarize then what we've done so far in dealing with the right term, referring to the two parts detailed here in B, the part dealing with simplifying the numerator and the part dealing with simplifying the denominator, we got that:

$1^4-0.5 =1-\frac{1}{2}=\frac{1}{2}$Where we used parentheses to emphasize the main fraction line,

Keep in mind that division is actually multiplication by the reciprocal number, and also that the reciprocal of a fraction is obtained by swapping the numerator and denominator, meaning:

$\frac{\frac{3}{2^3}-\big(1-\frac{1}{2}\big)^3}{1^4-0.5} = \frac{ \big(\frac{1}{4}\big)}{\big(\frac{1}{2}\big) }$We replace the division operation in the fraction with the multiplication by the reciprocal fraction that was detailed verbally earlier, we'll apply this to the expression we got in the last stage:

$\frac{\big(\frac{z}{w}\big)}{\big(\frac{x}{y}\big)}=\frac{z}{w}\cdot\frac{y}{x}$From here we'll continue as usual and perform the multiplication operation between the fractions, where we remember that when multiplying between fractions we multiply numerator by numerator and denominator by denominator and keep the original fraction line:

$\frac{\big(\frac{z}{w}\big)}{\textcolor{blue}{\big(\frac{x}{y}\big)}}=\frac{z}{w}\cdot\textcolor{blue}{\frac{y}{x}} \\ \downarrow\\ \frac{ \big(\frac{1}{4}\big)}{\textcolor{blue}{\big(\frac{1}{2}\big) }} =\frac{1}{4}\cdot \textcolor{blue}{\frac{2}{1} }$We reduce the fraction that we got (by dividing).

Let's summarize then what we've done so far in dealing with the right term, we got that:

$\frac{1}{4}\cdot\frac{2}{1} =\frac{1\cdot2}{4\cdot1}=\frac{\not{2}}{\not{4}}=\frac{1}{2}$And we've finished dealing with the right term,

Now let's return to the original problem and substitute the results of simplifying the expressions on the left and right that were detailed in A and B respectively:

$\frac{\frac{3}{2^3}-\big(1-\frac{1}{2}\big)^3}{1^4-0.5} = \frac{ \big(\frac{1}{4}\big)}{\big(\frac{1}{2}\big) } =\frac{1}{4}\cdot \frac{2}{1}=\frac{1}{2}$In order to determine which expression is larger we can represent the left expression as a fraction with denominator 2, this by expanding it, however since here the left expression is clearly larger than the number 1, while the right expression is smaller than the number 1 (we know this because it's a fraction with a smaller numerator than denominator) meaning that:$\frac{3^2\cdot(8-2\cdot3)^3}{(5^2\cdot3-72)\cdot\sqrt{16}}\text{ }_{\textcolor{red}{—}\text{ }}\frac{\frac{3}{2^3}-\big(1-\frac{1}{2}\big)^3}{1^4-0.5} \\ \downarrow\\ 6\text{ }_{\textcolor{red}{—}\text{ }}\frac{1}{2}$And therefore it's certainly true that:

$6>1>\frac{1}{2}$Meaning that the correct answer is answer B.

Let's handle each expression separately, the expression on the left and the expression on the right:

a. Let's start with the expression on the left:

$\frac{5^3\cdot(3^2-\sqrt{81})+6^2:12}{\sqrt{9}}$Remember that exponents come before multiplication and division, which come before addition and subtraction, and parentheses come before everything,

Therefore, we'll start by simplifying the expression in parentheses in the denominator:

$\frac{5^3\cdot(3^2-\sqrt{81})+6^2:12}{\sqrt{9}} = \frac{5^3\cdot(9-9)+6^2:12}{\sqrt{9}} =\\ \frac{5^3\cdot0+6^2:12}{\sqrt{9}}$where in the first stage we calculated the numerical value of the terms inside the parentheses, meaning - we performed the root operation and the exponentiation of the other term in parentheses, then we performed the subtraction operation within the parentheses and simplified the resulting expression,

Let's continue simplifying the expression remembering that multiplying any number by 0 will always give the result 0, simultaneously let's calculate the result of the root in the denominator in the last expression we got: $\frac{5^3\cdot0+6^2:12}{\sqrt{9}} = \frac{0+6^2:12}{3} =\\ \frac{6^2:12}{3}$Let's continue and calculate the term raised to the eighth power and perform the division operation in the denominator and finally perform the main fraction operation:

$\frac{6^2:12}{3}= \frac{36:12}{3}= \frac{3}{3}=1$where in the final stage we remembered that dividing any number by itself will always give the result 1,

We have finished handling the expression on the left,

Let's summarize the simplification steps:

$\frac{5^3\cdot(3^2-\sqrt{81})+6^2:12}{\sqrt{9}} = \frac{5^3\cdot0+6^2:12}{\sqrt{9}} =\\ \frac{6^2:12}{3}=1$

b. Let's continue and handle the expression on the right:

$\frac{(10^2-5)\cdot(9^2-81)+9^2:27}{\sqrt{9}}$First, let's simplify the expressions inside the parentheses in the denominator by calculating the terms with exponents and then performing the subtraction operation:

$\frac{(10^2-5)\cdot(9^2-81)+9^2:27}{\sqrt{9}} = \frac{(100-5)\cdot(81-81)+9^2:27}{\sqrt{9}} =\\ \frac{95\cdot0+9^2:27}{\sqrt{9}}$Again, let's remember that multiplying any number by 0 will always give the result 0, and simultaneously let's calculate the value of the term raised to the eighth power and the value of the root in its denominator:

$\frac{95\cdot0+9^2:27}{\sqrt{9}} = \frac{0+81:27}{3} =\\ \frac{81:27}{3}$Let's continue and perform the division operation in the denominator, and then calculate the value of the fraction:

$\frac{81:27}{3} = \frac{3}{3} =1$We have thus completed handling the expression on the right as well,

Let's summarize the simplification steps:

$\frac{(10^2-5)\cdot(9^2-81)+9^2:27}{\sqrt{9}} = \frac{95\cdot0+9^2:27}{\sqrt{9}}=\\ \frac{81:27}{3} = 1$

Now let's return to the original problem and substitute the results of simplifying the expressions from the left and right that were detailed in a and b and answer what was asked:

$\frac{5^3\cdot(3^2-\sqrt{81})+6^2:12}{\sqrt{9}}\text{ }{\textcolor{red}{—}\text{ }}\frac{(10^2-5)\cdot(9^2-81)+9^2:27}{\sqrt{9}} \\ \downarrow\\ 1\text{ }_{\textcolor{red}{—}\text{ }}1$We got, of course, that there is equality between the expression on the left and the expression on the right:$1\text{ }{\textcolor{red}{=}\text{ }}1$Therefore, the correct answer is answer c.