Solve for the Missing Number: Complex Fraction Equation with Square Roots and Powers

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