Complex Fraction Comparison: Finding the Missing Sign Between (5³×(3²-√81)+6²÷12)/√9 and Similar Expression

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.