Powers and Roots: Complete the missing numbers

Let's simplify the direct calculation of the left side of the equation:

$6^1+1^6+\sqrt{81}=\textcolor{red}{☐}^2 \\ 6+1+9=\textcolor{red}{☐}^2\\ 16=\textcolor{red}{☐}^2\\$When we calculated the numerical value of the term with the exponent and the term with the root, and remembered that raising the number 1 to any power will always give the result 1,

Now let's examine the equation we got, on the left side we have the number 16 and on the right side we have a number (unknown) raised to the second power,

So we ask the question: "What number do we need to square to get the number 16?"

And the answer to that is of course - the number 4,

Therefore:

$16=\textcolor{red}{4}^2$However, since we're dealing with an even power (power of 2), we must also consider the negative possibility,

Meaning it also holds true that:

$16=\textcolor{red}{(-4)}^2$

Therefore, the correct answer is answer C.

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 simplify separately each of the expressions:

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 since no mathematical operation is applied to them, therefore we can remove them, while writing the division operation as a fraction:

$(9^2-2\sqrt{81})+4^3:2^3=\\ 9^2-2\sqrt{81}+\frac{4^3}{2^3}$Next, we'll notice 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:

$\frac{a^m}{a^n}=a^{m-n}$Let's also remember the power rule for power of a power:

$(a^m)^n=a^{m\cdot n}$And we'll 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 complete expression and complete its calculation by substituting what we've gotten 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:

$8^2+2\sqrt{64}-3^2$And we'll 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:

$71\text{ }\textcolor{red}{{=\text{ }}}71$Therefore the correct answer is answer c.

Let's recall two laws of exponents:

a. The definition of root as an exponent:

$\sqrt[n]{a}=a^{\frac{1}{n}}$b. The law of exponents for exponents within parentheses where terms are multiplied, but in the opposite direction:

$x^n\cdot y^n =(x\cdot y)^n$Unlike previous questions where we converted the square root to its corresponding half power according to the definition of root as an exponent, in this problem we will not make this conversion but rather understand and internalize that according to the definition of root as an exponent, the root is an exponent in every way and therefore all laws of exponents apply to it, particularly the law of exponents mentioned in b',

We'll start then by simplifying the expression in the right parentheses (as written there, actually this is the denominator of the fraction equivalent to the division operation, we'll emphasize this later), we'll simplify this expression while applying the above understanding and using the law of exponents mentioned in b':

$(\sqrt{5}\cdot\sqrt{2})^2=(\sqrt{5\cdot2})^2=(\sqrt{10})^2$where in the first stage we applied the above understanding and treated the square root (which is actually a half power) as an exponent in every way and applied the law of exponents mentioned in b', we applied this law in the direction specified there by noting that the two multiplication terms have the same exponent (half power - of the square root) and therefore we could combine them together as multiplication between the bases under the same root (same exponent), in the next stage we simplified the expression under the root,

Now we'll remember that square root and square power are inverse operations and therefore cancel each other out, therefore:

$(\sqrt{10})^2 =10$Now we'll return to the complete expression on the right side of the equation and insert this information, in the next stage we'll write the division operation as a fraction:

$\sqrt{\textcolor{red}{☐}}=(2^5+18):(\sqrt{5}\cdot\sqrt{2})^2 \\ \downarrow\\ \sqrt{\textcolor{red}{☐}}=(2^5+18):10\\ \sqrt{\textcolor{red}{☐}}=\frac{2^5+18}{10}$$$Note that the division operation acts on the parentheses and therefore refers to them in their entirety, therefore the entire expression in parentheses enters the numerator of the fraction,

Let's continue and simplify the expression in the numerator of the fraction on the right side:

$\sqrt{\textcolor{red}{☐}}=\frac{2^5+18}{10} \\ \sqrt{\textcolor{red}{☐}}=\frac{32+18}{10} \\ \sqrt{\textcolor{red}{☐}}=\frac{\not{50}}{\not{10}} \\ \sqrt{\textcolor{red}{☐}}=5$where in the first stage we calculated the numerical value of the term with the exponent in the numerator of the fraction and in the next stage we performed the addition operation in the numerator, in the final stage we reduced the resulting fraction (performing the division operation in fact),

Let's summarize the solution steps so far, we got that:

$\sqrt{\textcolor{red}{☐}}=(2^5+18):(\sqrt{5}\cdot\sqrt{2})^2 \\ \sqrt{\textcolor{red}{☐}}=(2^5+18):(\sqrt{10})^2 \\ \sqrt{\textcolor{red}{☐}}=(2^5+18):10\\ \downarrow\\ \sqrt{\textcolor{red}{☐}}=\frac{2^5+18}{10} \\ \sqrt{\textcolor{red}{☐}}=\frac{\not{50}}{\not{10}} \\ \sqrt{\textcolor{red}{☐}}=5$

Let's examine now the expression we got:

On the right side we got the number 5, and on the left side there's a (unknown) number that's under a square root,

In other words - we need to answer the question - "The square root of which number is 5?"

The answer to this question is of course the number: 25, it's easy to verify this by calculating its square root which indeed gives the result 5,

Meaning - for the equation to be true, under the root must be the number 25:

$\sqrt{\textcolor{red}{25}}=5$Therefore the correct answer is answer d.

First, let's calculate by direct computation the values of all terms in the equation:

$7^1+3^4=4^3+\sqrt{\textcolor{red}{☐}}+2^3 \\ 7+81=64+\sqrt{\textcolor{red}{☐}}+8$remembering that raising any number to the power of 1 will always give the number itself,

Next, we'll remember that this is an equation and therefore we can move terms from one side to the other, doing so while remembering that when a term moves sides it changes its sign:

$7+81=64+\sqrt{\textcolor{red}{☐}}+8 \\ 7+81-64-8=\sqrt{\textcolor{red}{☐}}\\ 16 =\sqrt{\textcolor{red}{☐}}$In the final step, we simplified the left side of the equation by combining like terms,

Now let's examine the equation we got:

On the left side we have the number 16 and on the right side we have a number (which is unknown) under a square root,

Therefore we ask the question: "The square root of which number is 16?"

We can answer this question by guessing and checking the square roots of different numbers using a calculator, but a better way is to remember that square root and squaring are inverse operations and therefore:

$16=\sqrt{16^2}$Therefore the answer to the above question is of course the number:

$16^2$Let's calculate the numerical value of this term:

$16^2=256$Therefore the answer to the above question, meaning - the unknown number under the square root in the problem, is the number 256:

$16 =\sqrt{\textcolor{red}{256}}$Therefore the correct answer is answer B.

To solve the problem, we need to evaluate both sides of the equation step by step and determine which number should replace the missing value to maintain equality. We'll start by evaluating the left side and simplifying the right side to find the missing number.

Let's examine the left side of the equation:
$(2^3+5^2)-9^2:3^2-\sqrt{100}$

• First, calculate $2^3$ which is $8$.

• Next, calculate $5^2$ which is $25$.

• Add these results together: $8 + 25 = 33$.

• Calculate $9^2$ which is $81$.

• Calculate $3^2$ which is $9$.

• Divide $81$ by $9$:$\ 81 \div 9 = 9$.

• Subtract $9$ from $33$: $33 - 9 = 24$.

• Calculate $\sqrt{100}$ which is $10$.

• Subtract $10$ from $24$:$24 - 10 = 14$.

Now, let's simplify the right side of the equation:
$(6\cdot5-\textcolor{red}{☐}^2)^2-\sqrt{25}-6$

• Calculate $6\cdot5$ which is $30$.

• The expression becomes: $(30-\textcolor{red}{☐}^2)^2-\sqrt{25}-6$.

• We know from the original problem statement that the complete expression must equal $14$.

• Calculate $\sqrt{25}$, which is $5$.

• The equation: $(30-\textcolor{red}{☐}^2)^2-5-6=14$.

• Simplify further:
  $((30-\textcolor{red}{☐}^2)^2=25$

• Taking square roots of both sides: $30-\textcolor{red}{☐}^2=5$ or $-(30-\textcolor{red}{☐}^2)=5$

• Solving for $\textcolor{red}{☐}$:

  • $30-\textcolor{red}{☐}^2=5\rightarrow \textcolor{red}{☐}^2=25\rightarrow \textcolor{red}{☐}=5$

  • Alternatively, for the negative case: $-(30-\textcolor{red}{☐}^2)=5\rightarrow 30-\textcolor{red}{☐}^2=-5$ does not result in a real number solution.

Therefore, $\textcolor{red}{☐}$ must be $5$ to balance the equation

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.

To solve the equation $\frac{(6^2-4^2):\sqrt{25}}{\sqrt{4}}=\frac{\sqrt{36}-\sqrt{49}:7}{5}+\textcolor{red}{☐}^{100}$, we need to simplify both sides step by step.

Let's start with the left-hand side (LHS):

• Calculate the powers: $6^2 = 36$ and $4^2 = 16$.
• Subtract the results: $36 - 16 = 20$.
• Calculate the square root: $\sqrt{25} = 5$.
• Perform the division: $20 : 5 = 4$.
• Calculate the square root in the denominator: $\sqrt{4} = 2$.
• Complete the division to simplify: $\frac{4}{2} = 2$.

So, the LHS simplifies to $2$.

Now, let's simplify the right-hand side (RHS):

• Calculate the square roots: $\sqrt{36} = 6$ and $\sqrt{49} = 7$.
• Perform the division inside the expression: $7:7 = 1$.
• Subtract the results: $6 - 1 = 5$.
• Divide by 5: $\frac{5}{5} = 1$.

So, the RHS simplifies to $1 + \textcolor{red}{☐}^{100}$.

This gives us the equation:

• $2 = 1 + \textcolor{red}{☐}^{100}$
• Solve for the missing number: $\textcolor{red}{☐}^{100} = 1$.

We know that $x^{100} = 1$ has two solutions for any real numbers: $x = 1$ and $x = -1$ because both $1^{100}$ and $(-1)^{100} = 1$ hold true.

Thus, the missing number is $1,\hspace{4pt}-1$.

Let's simplify the equation, dealing with the fractions in both sides separately:

A. We'll start with the fraction on the left side:

$\frac{(-1)^8-(-3)\cdot\sqrt{16}+3}{6^2:(-3)^2-2^2\cdot2}$Let's simplify this fraction while remembering the order of operations, meaning that exponents come before division and multiplication, which come before addition and subtraction, and parentheses come before everything,

Additionally, we'll remember that an exponent is multiplying a number by itself and therefore raising any number (even a negative number) to an even power will give a positive result, and this is because negative one multiplied by negative one gives the result of one,

In particular, in this problem:

$(-1)^8=1\\ (-3)^2=9$Let's return to the fraction and apply this, first we'll deal with the numerator of the fraction where we'll calculate the numerical value of the square root in the second term from the left and the value of the term with the exponent:

$\frac{(-1)^8-(-3)\cdot\sqrt{16}+3}{6^2:(-3)^2-2^2\cdot2}=\\ \frac{1-(-3)\cdot4+3}{6^2:(-3)^2-2^2\cdot2}$Now we'll remember that according to the multiplication rules, multiplying a negative number by a negative number will give a positive result, we'll apply this to the numerator of the fraction in question, then we'll perform the multiplication in the second term from the left and simplify the expression in the numerator of the fraction:

$\frac{1-(-3)\cdot4+3}{6^2:(-3)^2-2^2\cdot2} =\\ \frac{1+3\cdot4+3}{6^2:(-3)^2-2^2\cdot2} =\\ \frac{1+12+3}{6^2:(-3)^2-2^2\cdot2}=\\ \frac{16}{6^2:(-3)^2-2^2\cdot2}$We'll continue and simplify the denominator of the fraction, we'll start by calculating the values of the terms with exponents, this we'll do, again, using what was said before (regarding the even exponent), then we'll perform the division operation in the first term from the left and the multiplication operation in the second term from the left, and finally we'll perform the subtraction operation in the denominator of the fraction:

$\frac{16}{6^2:(-3)^2-2^2\cdot2} =\\ \frac{16}{36:9-4\cdot2}=\\ \frac{16}{4-8}=\\ \frac{16}{-4}=\\ -4$In the last stage we performed the division operation of the fraction, this while we remember that according to the division rules (which are identical to the multiplication rules in this context) dividing a positive number by a negative number will give a negative result,

We have finished simplifying the fraction on the left side, let's summarize this simplification:

$\frac{(-1)^8-(-3)\cdot\sqrt{16}+3}{6^2:(-3)^2-2^2\cdot2}=\\ \frac{1-(-3)\cdot4+3}{6^2:(-3)^2-2^2\cdot2} =\\\ \frac{16}{6^2:(-3)^2-2^2\cdot2} =\\ \frac{16}{36:9-4\cdot2}=\\ \frac{16}{-4}=\\ -4$

B. Let's simplify the fraction on the right side:

$\frac{(-2)^3+\textcolor{red}{☐}^2-1}{-(-2)^2}$For convenience, let's name the number we're looking for and define it as x:

$\textcolor{red}{☐}=x\\ \downarrow\\ \textcolor{red}{☐}^2=x^2$Let's substitute this in the fraction in question:

$\frac{(-2)^3+x^2-1}{-(-2)^2}$So let's start simplifying the fraction, again we'll use the fact that raising any number to an even power will give a positive result, but we'll also remember the fact that raising a negative number to an odd power will give a negative result, in particular in the problem:

$(-2)^2=4\\ (-2)^3=-8$Let's apply this to the fraction in question:

$\frac{(-2)^3+x^2-1}{-(-2)^2} =\\ \frac{-8+x^2-1}{-4}$Now let's combine like terms in the numerator of the fraction:

$\frac{-8+x^2-1}{-4}=\\ \frac{-9+x^2}{-4}$We have finished simplifying the fraction on the right side of the given equation, let's summarize this simplification: