Let's solve the problems from the image step by step:

1. Найдите $9^{a-1}$, если $3^{a-1} = 4$

We know that $9^{a-1}$ can be rewritten as:

$9^{a-1} = (3^2)^{a-1} = 3^{2(a-1)}$

We are also given that $3^{a-1} = 4$. So,

$9^{a-1} = 3^{2(a-1)} = (3^{a-1})^2 = 4^2 = 16$

Thus, $9^{a-1} = 16$.

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2. Найдите $5^{x+1}$, если $25^x = 81$

First, rewrite $25^x$ as $(5^2)^x = 5^{2x}$, so the equation becomes:

$5^{2x} = 81$

Now, take the square root of both sides:

$5^x = \sqrt{81} = 9$

Next, we need to find $5^{x+1}$:

$5^{x+1} = 5^x \cdot 5 = 9 \cdot 5 = 45$

Thus, $5^{x+1} = 45$.

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3. Упростите выражение $\frac{10^{m+1} - 10^{m-1}}{10^{m+1} + 10^m}$

Factor both the numerator and the denominator.

In the numerator, factor out $10^{m-1}$:

$10^{m+1} - 10^{m-1} = 10^{m-1}(10^2 - 1) = 10^{m-1}(100 - 1) = 10^{m-1}(99)$

In the denominator, factor out $10^m$:

$10^{m+1} + 10^m = 10^m(10 + 1) = 10^m(11)$

Now the expression becomes:

$\frac{10^{m-1} \cdot 99}{10^m \cdot 11}$

Simplify by canceling $10^{m-1}$:

$\frac{99}{11 \cdot 10} = \frac{9}{10}$

Thus, the simplified expression is $\frac{9}{10}$.

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4. Найдите $\frac{9^{m+1}}{3^{2m-2}}$, если $3^m = x$

First, rewrite $9^{m+1}$ in terms of base 3:

$9^{m+1} = (3^2)^{m+1} = 3^{2(m+1)} = 3^{2m+2}$

Now, the expression becomes:

$\frac{3^{2m+2}}{3^{2m-2}} = 3^{(2m+2) - (2m-2)} = 3^{4}$

Thus, the result is:

$3^4 = 81$

So, $\frac{9^{m+1}}{3^{2m-2}} = 81$.

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5. Найдите $9^n + 9^{-n}$, если $3^n - 3^{-n} = 3$

First, express everything in terms of base 3. We know that $9^n = (3^2)^n = 3^{2n}$ and $9^{-n} = 3^{-2n}$.

Given $3^n - 3^{-n} = 3$, let’s denote:

$y = 3^n$

This means the given equation becomes:

$y - \frac{1}{y} = 3$

Multiply both sides by $y$:

$y^2 - 1 = 3y$

Rearranging gives:

$y^2 - 3y - 1 = 0$

Solve this quadratic equation using the quadratic formula:

$y = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-1)}}{2(1)} = \frac{3 \pm \sqrt{9 + 4}}{2} = \frac{3 \pm \sqrt{13}}{2}$

Now, we need to find $9^n + 9^{-n}$, which is $y^2 + \frac{1}{y^2}$. From the identity:

$y^2 + \frac{1}{y^2} = (y - \frac{1}{y})^2 + 2$

We know that $y - \frac{1}{y} = 3$, so:

$y^2 + \frac{1}{y^2} = 3^2 + 2 = 9 + 2 = 11$

Thus, $9^n + 9^{-n} = 11$.

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Do you need further clarification on any of the steps? Here are 5 related problems:

1. Solve $5^{2x} = 64$ for $x$.
2. Simplify $\frac{10^{n+2} - 10^n}{10^{n+1} + 10^{n-1}}$.
3. If $4^x = 16$, find $2^{2x+1}$.
4. Simplify $\frac{27^{m+1}}{3^{3m-3}}$ in terms of base 3.
5. Find $3^{2n} + 3^{-2n}$ if $3^n - 3^{-n} = 2$.

Tip: When dealing with exponents, always look for opportunities to rewrite terms with the same base to simplify the expression.