https://file.aiquickdraw.com/mathbot/file/17284523702150egsjcw4.jpg

Let's go through the problems one by one:

### 1. Solve the equation:

1) \(\left(\frac{1}{5}\right)^{2 - 3x} = 25\)

We can express \(25\) as \(5^2\), so the equation becomes:
\[
\left(\frac{1}{5}\right)^{2 - 3x} = \left(5\right)^2 = \left(\frac{1}{5}\right)^{-2}
\]
Equating the exponents:
\[
2 - 3x = -2
\]
Solving for \(x\):
\[
-3x = -2 - 2 \quad \Rightarrow \quad -3x = -4 \quad \Rightarrow \quad x = \frac{4}{3}
\]

2) \(4^x + 2^x - 20 = 0\)

Let’s set \(y = 2^x\). Then \(4^x = (2^2)^x = y^2\), so the equation becomes:
\[
y^2 + y - 20 = 0
\]
This is a quadratic equation. Solving it using the quadratic formula:
\[
y = \frac{-1 \pm \sqrt{1^2 - 4(1)(-20)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 80}}{2} = \frac{-1 \pm \sqrt{81}}{2} = \frac{-1 \pm 9}{2}
\]
Thus:
\[
y = \frac{-1 + 9}{2} = 4 \quad \text{or} \quad y = \frac{-1 - 9}{2} = -5
\]
Since \(y = 2^x\) must be positive, \(y = 4\), so \(2^x = 4\) gives \(x = 2\).

### 2. Solve the inequality:

\[
\left(\frac{3}{4}\right)^x > \frac{1}{3}
\]

Taking the logarithm of both sides (since \(\frac{3}{4} < 1\), the inequality will reverse):
\[
x \ln\left(\frac{3}{4}\right) < \ln\left(\frac{1}{3}\right)
\]
Since \(\ln\left(\frac{3}{4}\right)\) is negative, we divide by it and reverse the inequality:
\[
x > \frac{\ln\left(\frac{1}{3}\right)}{\ln\left(\frac{3}{4}\right)}
\]
Calculating the result gives the value for \(x\).

### 3. Solve the system of equations:

\[
\begin{cases}
x - y = 4 \\
5x + y = 25
\end{cases}
\]

Adding the two equations to eliminate \(y\):
\[
(x - y) + (5x + y) = 4 + 25 \quad \Rightarrow \quad 6x = 29 \quad \Rightarrow \quad x = \frac{29}{6}
\]
Substituting \(x = \frac{29}{6}\) into the first equation:
\[
\frac{29}{6} - y = 4 \quad \Rightarrow \quad y = \frac{29}{6} - 4 = \frac{29}{6} - \frac{24}{6} = \frac{5}{6}
\]

Thus, \(x = \frac{29}{6}\) and \(y = \frac{5}{6}\).

### 4. Solve the inequality:

1) \(\left(\sqrt{5}\right)^{x - 6} < \frac{1}{5}\)

Express \(\frac{1}{5}\) as \(5^{-1}\), and \(\sqrt{5}\) as \(5^{1/2}\):
\[
\left(5^{1/2}\right)^{x - 6} < 5^{-1}
\]
Simplifying the exponent:
\[
5^{\frac{x - 6}{2}} < 5^{-1}
\]
Equating the exponents:
\[
\frac{x - 6}{2} < -1 \quad \Rightarrow \quad x - 6 < -2 \quad \Rightarrow \quad x < 4
\]

2) \(\left(\frac{2}{13}\right)^{x^2 - 1} \geq 1\)

Since any base raised to 0 equals 1, we solve:
\[
x^2 - 1 \geq 0 \quad \Rightarrow \quad x^2 \geq 1 \quad \Rightarrow \quad x \geq 1 \quad \text{or} \quad x \leq -1
\]

### 5. Solve the equation:

\[
7^{x+1} + 3 \cdot 7^x = 2x^5 + 3 \cdot 2x
\]
This equation seems to involve complex exponential and polynomial terms, and would require further methods like substitution or numerical solving depending on the given terms.

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Let me know if you'd like more details on any specific step, or if you have any questions!

Here are 5 related problems:
1. Solve the equation \(\left(\frac{1}{4}\right)^{x-2} = 16\).
2. Solve the inequality \(\left(\frac{5}{6}\right)^x < \frac{2}{5}\).
3. Solve the system of equations \(\begin{cases} 2x + 3y = 7 \\ 4x - y = 5 \end{cases}\).
4. Solve the inequality \(\left(2^{x-3}\right) > \frac{1}{8}\).
5. Solve the equation \(5^{x+2} + 25^{x-1} = 30\).

**Tip:** When solving systems of equations, elimination or substitution are both powerful methods! Choose the one that simplifies the problem faster.

Решить уравнение и неравенство: 1) (1/5)^(2-3x) = 25, 2) 4^x + 2^x - 20 = 0, неравенство (3/4)^x > 1/3, систему уравнений {x-y=4, 5x+y=25}, неравенство (√5)^(x-6) < 1/5 и (2/13)^(x^2-1) >= 1, уравнение 7^(x+1) + 3 * 7^x = 2x^5 + 3 * 2x.

Step-by-step solutions for solving exponential equations like (1/5)^(2-3x) = 25, quadratic equations such as 4^x + 2^x - 20 = 0, as well as inequalities and systems of linear equations. Detailed explanations cover key algebraic concepts, including logarithmic properties and quadratic formula application. Ideal for students in Grades 9-11.

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