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Let's go through the problems step by step. I'll solve each equation for you.

### 1. \( 5 + \frac{3}{x} = \frac{3}{4x^2} \)

**Solution:**

We want to clear the fractions by multiplying through by the least common denominator (LCD). The LCD here is \( 4x^2 \).

Multiplying everything by \( 4x^2 \):

\[
4x^2(5) + 4x^2\left(\frac{3}{x}\right) = 4x^2\left(\frac{3}{4x^2}\right)
\]

This simplifies to:

\[
20x^2 + 12x = 3
\]

Now, rearrange the equation to form a quadratic:

\[
20x^2 + 12x - 3 = 0
\]

Solve this quadratic using the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Where \( a = 20 \), \( b = 12 \), and \( c = -3 \).

Now substitute the values into the formula:

\[
x = \frac{-12 \pm \sqrt{12^2 - 4(20)(-3)}}{2(20)}
\]
\[
x = \frac{-12 \pm \sqrt{144 + 240}}{40}
\]
\[
x = \frac{-12 \pm \sqrt{384}}{40}
\]

\[
x = \frac{-12 \pm 19.6}{40}
\]

So, the two possible solutions are:

\[
x_1 = \frac{-12 + 19.6}{40} = \frac{7.6}{40} = 0.19
\]
\[
x_2 = \frac{-12 - 19.6}{40} = \frac{-31.6}{40} = -0.79
\]

### 2. \( 1 + \frac{2}{x} = 0 \)

**Solution:**

Rearrange the equation:

\[
\frac{2}{x} = -1
\]

Multiply both sides by \( x \):

\[
2 = -x
\]

Solve for \( x \):

\[
x = -2
\]

### 3. \( \frac{4x}{3x - 5x} + \frac{1}{x} = 7 \left( \frac{8 - x}{5x - x^2} \right) \)

This one appears to have a more complex structure. To proceed, I'll need to clarify some aspects. Specifically, the first term \( 4x/(3x - 5x) \) simplifies to zero because \( 3x - 5x = -2x \). Could you confirm if this term is intended to be different, or should I continue solving with it? Let me know!

Solve the following rational equations: 1) 5 + 3/x = 3/4x^2; 2) 1 + 2/x = 0; 3) (4x)/(3x-5x) + 1/x = 7 * (8-x)/(5x - x^2); 4) 5/(x^2-1) = 2/(x+1); 5) 1 + 1/5x = (8-2x)/-10x

Learn how to solve rational equations involving fractions and quadratic expressions. This problem set covers key algebraic concepts and requires solving equations through clearing denominators and applying the quadratic formula. Ideal for students in Grades 9-11.

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