Answer :

Sure, let's solve each equation step by step.

### Equation 1:
[tex]\[ \frac{2 x}{x+1} + \frac{5}{2 x} = 2 \][/tex]

First, let's find a common denominator, which is [tex]\(2x(x+1)\)[/tex]:

[tex]\[ \frac{(2x)^2 + 5(x + 1)}{2x(x + 1)} = 2 \][/tex]

Simplify the numerator:

[tex]\[ \frac{4x^2 + 5x + 5}{2x(x + 1)} = 2 \][/tex]

Cross-multiply to clear the fraction:

[tex]\[ 4x^2 + 5x + 5 = 4x(x + 1) \][/tex]

Expand the right-hand side:

[tex]\[ 4x^2 + 5x + 5 = 4x^2 + 4x \][/tex]

Subtract [tex]\(4x^2\)[/tex] from both sides:

[tex]\[ 5x + 5 = 4x \][/tex]

Subtract [tex]\(4x\)[/tex] from both sides:

[tex]\[ x + 5 = 0 \][/tex]

Therefore, solving for [tex]\(x\)[/tex]:

[tex]\[ x = -5 \][/tex]

So the solution to the first equation is:

[tex]\[ x = -5 \][/tex]

### Equation 2:
[tex]\[ \frac{5}{v} + \frac{5}{v + 10} = \frac{4}{3} \][/tex]

Find a common denominator, which is [tex]\(v(v + 10)\)[/tex]:

[tex]\[ \frac{5(v + 10) + 5v}{v(v + 10)} = \frac{4}{3} \][/tex]

Simplify the numerator:

[tex]\[ \frac{5v + 50 + 5v}{v(v + 10)} = \frac{4}{3} \][/tex]

Combine like terms:

[tex]\[ \frac{10v + 50}{v(v + 10)} = \frac{4}{3} \][/tex]

Cross-multiply to clear the fraction:

[tex]\[ 3(10v + 50) = 4v(v + 10) \][/tex]

Expand both sides:

[tex]\[ 30v + 150 = 4v^2 + 40v \][/tex]

Rearrange to form a quadratic equation:

[tex]\[ 4v^2 + 40v - 30v - 150 = 0 \][/tex]
[tex]\[ 4v^2 + 10v - 150 = 0 \][/tex]

Divide the entire equation by 2 to simplify:

[tex]\[ 2v^2 + 5v - 75 = 0 \][/tex]

Factor the quadratic equation:

[tex]\[ (2v + 15)(v - 5) = 0 \][/tex]

So, the solutions for [tex]\(v\)[/tex] are:

[tex]\[ 2v + 15 = 0 \implies v = -\frac{15}{2} = -7.5 \][/tex]
[tex]\[ v - 5 = 0 \implies v = 5 \][/tex]

So the solutions to the second equation are:

[tex]\[ v = -7.5 \quad\text{and}\quad v = 5 \][/tex]

### Equation 3:
[tex]\[ \frac{x^2 - 10}{x - 1} = \frac{-14 - 5x}{x - 1} \][/tex]

Since the denominators are the same, we can equate the numerators directly:

[tex]\[ x^2 - 10 = -14 - 5x \][/tex]

Rearrange to form a quadratic equation:

[tex]\[ x^2 + 5x - 10 + 14 = 0 \][/tex]
[tex]\[ x^2 + 5x + 4 = 0 \][/tex]

Factor the quadratic equation:

[tex]\[ (x + 4)(x + 1) = 0 \][/tex]

So, the solutions for [tex]\(x\)[/tex] are:

[tex]\[ x + 4 = 0 \implies x = -4 \][/tex]
[tex]\[ x + 1 = 0 \implies x = -1 \][/tex]

So the solutions to the third equation are:

[tex]\[ x = -4 \quad\text{and}\quad x = -1 \][/tex]

### Summary:
1. [tex]\( x = -5 \)[/tex]
2. [tex]\( v = -7.5 \)[/tex] and [tex]\( v = 5 \)[/tex]
3. [tex]\( x = -4 \)[/tex] and [tex]\( x = -1 \)[/tex]

Hence, the complete set of solutions is:
[tex]\[ x = -5 \][/tex]
[tex]\[ v = -7.5, v = 5 \][/tex]
[tex]\[ x = -4, x = -1 \][/tex]