Complex Fraction Comparison: Find the Missing Sign Between √16·(6²-21)/(64-54):2 and 9/√81-(1-2/3)²/(16:16)

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.