Powers and Roots: Complete the missing numbers

Fill in the missing sign:

^{$$$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$}

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.

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Fill in the missing sign:

$$$\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}$

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.

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Mark the correct sign:

$$$\frac{3^2\cdot(8-2\cdot3)^3}{(5^2\cdot3-72)\cdot\sqrt{16}}\text{ }_{\textcolor{red}{—}\text{ }}\frac{\frac{3}{2^3}-(1-\frac{1}{2})^3}{1^4-0.5}$

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.

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Indicate the missing number:

$(5^2+3^2):(\sqrt{16}\cdot\sqrt{9}+1+2^3:2)=\frac{(6^2-\sqrt{16}):\textcolor{red}{☐}^3}{\sqrt{4}}$

Let's simplify the expressions on both sides of the equation separately:

A. Let's start with the expression on the left side:

$(5^2+3^2):(\sqrt{16}\cdot\sqrt{9}+1+2^3:2)$Let's recall the order of operations in which powers take precedence over multiplication and division, and multiplication and division take precedence over addition and subtraction,

so let's start by simplifying the expressions inside the parentheses, where we will first calculate the values of the terms in the power and in the roots (which are powers of everything), then we will calculate the values of the products and results of the division operations, and finally we will perform the addition operations in the parentheses:

$(5^2+3^2):(\sqrt{16}\cdot\sqrt{9}+1+2^3:2) =\\ (25+9):(4\cdot3+1+8:2) =\\ 34:(12+1+4)=\\ 34:17=\\ 2$In the last step we calculated the result of the division operation that remained (which was originally between the two main parentheses),

We have thus completed the simplification of the expression on the left side.

B. Let's continue simplifying the expression on the right side:

$\frac{(6^2-\sqrt{16}):\textcolor{red}{☐}^3}{\sqrt{4}}$For convenience of operations, let's call this unknown number we are looking for, let's define it as- x:

$\textcolor{red}{☐}=x\\ \downarrow\\ \textcolor{red}{☐}^3=x^3\\$and we will place it in the mentioned expression:

$\frac{(6^2-\sqrt{16}):\textcolor{red}{☐}^3}{\sqrt{4}} \\ \downarrow\\ \frac{(6^2-\sqrt{16}):x^3}{\sqrt{4}}$Let's continue simplifying the expression, let's start by calculating their numerical value of the terms in the power and in the root in the numerator, in parallel we will calculate the numerical value of the term in the root in the denominator, then we will calculate the result of the subtraction operation in the numerator:

$\frac{(6^2-\sqrt{16}):x^3}{\sqrt{4}}= \\ \frac{(36-4):x^3}{2} =\\ \frac{32:x^3}{2}$For convenience of solution for now, we will leave this expression in its current form and emphasize that we have completed the treatment of the expression on the right side.

Let's go back to the original equation and place in it the two simplification results we got for the terms on the left and right sides that were detailed in A and B, let's not forget that we also defined the number we are looking for as x:

$(5^2+3^2):(\sqrt{16}\cdot\sqrt{9}+1+2^3:2)=\frac{(6^2-\sqrt{16}):\textcolor{red}{☐}^3}{\sqrt{4}} \\ \downarrow\\ (5^2+3^2):(\sqrt{16}\cdot\sqrt{9}+1+2^3:2)=\frac{(6^2-\sqrt{16}):x^3}{\sqrt{4}} \\ \downarrow\\ 2= \frac{32:x^3}{2} \\ \frac{2}{1}= \frac{32:x^3}{2}$$$In the last step we used the fact that any number can be written as a number divided by 1, we did this as a preparation for the next step where we will solve the equation that was obtained by multiplying both sides by the common denominator,

Now let's simplify the equation we got, again we will multiply both sides by the common denominator, but this time the common denominator is the algebraic expression: $x^3$, this expression depends on the unknown we are looking for and therefore we must define a permitted range of definition, since multiplying the equation by 0 is forbidden (and in general division by 0 is also a forbidden operation, so from the beginning we can define this range of definition):

$\frac{4}{1}=\frac{32}{x^3} \hspace{8pt}\text{/}\cdot x^3\hspace{4pt};x\neq0\\ 4 x^3=32$Again- we will know how much to multiply each term by using the answer to the question:"By how much did we multiply the current denominator in order to get the common denominator?", we also did not forget to define the permitted range of definition (we got it by solving the inequality:

$x^3\neq0$)

Let's continue and isolate the unknown by dividing both sides of the equation by its coefficient:

$4 x^3=32 \hspace{8pt}\text{/}:4\\ x^3=\frac{\not{32}}{\not{4}}\\ x^3=8$When in the last step we reduced the fraction that was obtained on the right side as a result of the division operation,

Let's finish solving the equation by taking the cube root of both sides of the equation, the cube root which is an operation inverse to the cube power will cancel the cube power on the unknown on the left side of the equation:

$x^3=8 \hspace{8pt}\text{/}\sqrt[3]{\hspace{4pt}}\\ \sqrt[3]{x^3}=\sqrt[3]{8}\\ x=2$Note that since we took an odd-order root from both sides of the equation we only need to consider one solution (it is the solution we will get in the calculator when connecting the cube root of the number 8) as opposed to taking an even-order root, in which case we need to consider two possible solutions - a positive solution and a negative solution,

We have thus solved the equation, and we got that the number we are looking for, which we defined to be the unknown x (marked at the beginning of the problem in a red square) is the number 2,

So the correct answer is answer D.

2

Indicate the missing number:

$$$\frac{(-1)^8-(-3)\cdot\sqrt{16}+3}{6^2:(-3)^2-2^2\cdot2}=\frac{(-2)^3+\textcolor{red}{☐}^2-1}{-(-2)^2}$

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:

$\frac{(-2)^3+x^2-1}{-(-2)^2} =\\ \frac{-8+x^2-1}{-4} =\\ \frac{-9+x^2}{-4}$

Let's now return to the original equation and substitute the results of simplifying the fractions on the left and right sides detailed in A and B respectively, we won't forget the definition of the unknown x mentioned earlier:

$\frac{(-1)^8-(-3)\cdot\sqrt{16}+3}{6^2:(-3)^2-2^2\cdot2}=\frac{(-2)^3+\textcolor{red}{☐}^2-1}{-(-2)^2} \\ \downarrow\\ \frac{(-1)^8-(-3)\cdot\sqrt{16}+3}{6^2:(-3)^2-2^2\cdot2}=\frac{(-2)^3+x^2-1}{-(-2)^2} \\ \downarrow\\ -4= \frac{-9+x^2}{-4}$ Let's continue and solve the resulting equation, first we'll remember that any number can be represented as itself divided by the number 1:

$-4= \frac{-9+x^2}{-4}\\ \frac{-4}{1}= \frac{-9+x^2}{-4}$Then we'll multiply both sides of the equation by the common denominator, which is the number $-4$ and then we'll simplify the equation and isolate the unknown by moving terms and combining like terms:

$\frac{-4}{1}= \frac{-9+x^2}{-4} \hspace{8pt}\text{/}\cdot(-4) \\ (-4)\cdot(-4)=-9+x^2\\ 16=-9+x^2\\ 16+9=x^2\\ 25=x^2$In the first stage we multiplied by the common denominator in order to cancel the fraction line while we multiply each numerator by the number that is the answer to the question "By how much did we multiply the current denominator to get the common denominator?" ,

Let's return to what was asked, we are looking for the number x (in the original problem it is a red square) for which we get a true statement from the last equation, we can take the square root of both sides of the equation and get the two possible solutions (we'll remember that taking a square root in the context of solving an equation always involves taking into account two possibilities, positive and negative) but we can also use logic and remember that raising any number to an even power will give a positive result, meaning the above equation has two possible answers, a positive answer and a negative answer,

We'll add and ask: "What number did we raise to the second power to get the number 25?" a question to which the answer is of course the number 5 (or minus 5), and therefore the full answer is:

$x=5,\hspace{4pt}-5$ Each of these options is correct,

And therefore the correct answer is answer C.

$5,\hspace{4pt}-5$

Indicate the missing number:

$6^1+1^6+\sqrt{81}=\textcolor{red}{☐}^2$

$4,\hspace{4pt}-4$

Mark the appropriate sign:

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

$=$

Mark the appropriate sign:

^{$(10^2:\sqrt{16}-2^2\cdot6)^{100}\text{ }_{\textcolor{red}{—}\text{ }}(7:\sqrt{49})+3^2-2^3$}

<

Indicate the missing number:

$7^1+3^4=4^3+\sqrt{\textcolor{red}{☐}}+2^3$

256

Indicate the missing number:

^{$\sqrt{\textcolor{red}{☐}}=(2^5+18):(\sqrt{5}\cdot\sqrt{2})^2$}

25

Indicate the missing number:

^{$(2^3+5^2)-9^2:3^2-\sqrt{100}=(6\cdot5-\textcolor{red}{☐}^2)^2-\sqrt{25}-6$}

5

Fill in the missing sign:

^{$$$\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}}$}

$=$

Indicate the missing number:

$\frac{(6^2-4^2):\sqrt{25}}{\sqrt{4}}=\frac{\sqrt{36}-\sqrt{49}:7}{5}+\textcolor{red}{☐}^{100}$

$1,\hspace{4pt}-1$

Indicate the missing number:

$\sqrt{\frac{64}{10000}}+\frac{92}{10^2}=\textcolor{red}{☐}^{450}$

$1,\hspace{4pt}-1$