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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, \( 9^{a-1} = 16 \).

---

### 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, \( 5^{x+1} = 45 \).

---

### 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 \( \frac{9}{10} \).

---

### 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, \( \frac{9^{m+1}}{3^{2m-2}} = 81 \).

---

### 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, \( 9^n + 9^{-n} = 11 \).

---

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.

Learn how to solve exponential equations and simplify expressions with detailed steps. This problem set covers key mathematical concepts such as laws of exponents, simplifying algebraic expressions, and quadratic equations. Suitable for students in Grades 10-12.

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