Compare Complex Fractions: Evaluating (3²×(8-2×3)³) ÷ (5²×3-72)×√16

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.