Let's solve the given algebraic expressions step-by-step.

### Problem 19:
\[
\frac{x+1}{x-1} + \frac{x-1}{x+1} = \frac{4}{(x+1)(x-1)}
\]

First, find a common denominator:
\[
\frac{(x+1)^2 + (x-1)^2}{(x-1)(x+1)} = \frac{4}{(x+1)(x-1)}
\]

Simplify the numerator:
\[
(x+1)^2 + (x-1)^2 = x^2 + 2x + 1 + x^2 - 2x + 1 = 2x^2 + 2
\]

So, we have:
\[
\frac{2x^2 + 2}{(x-1)(x+1)} = \frac{4}{(x+1)(x-1)}
\]

Simplify:
\[
2(x^2 + 1) = 4 \implies x^2 + 1 = 2 \implies x^2 = 1 \implies x = \pm 1
\]

### Problem 20:
\[
\frac{2}{x-2} + \frac{3}{x-3} = \frac{3x-20}{(x-2)(x-3)}
\]

Find a common denominator:
\[
\frac{2(x-3) + 3(x-2)}{(x-2)(x-3)} = \frac{3x-20}{(x-2)(x-3)}
\]

Simplify the numerator:
\[
2(x-3) + 3(x-2) = 2x - 6 + 3x - 6 = 5x - 12
\]

So, we have:
\[
\frac{5x - 12}{(x-2)(x-3)} = \frac{3x-20}{(x-2)(x-3)}
\]

Equate the numerators:
\[
5x - 12 = 3x - 20 \implies 2x = -8 \implies x = -4
\]

### Problem 21:
\[
\frac{x-1}{x-2} + \frac{8}{(x-2)(x-4)} = \frac{x+3}{x-4}
\]

Find a common denominator:
\[
\frac{(x-1)(x-4) + 8}{(x-2)(x-4)} = \frac{x+3}{x-4}
\]

Simplify the numerator:
\[
(x-1)(x-4) + 8 = x^2 - 5x + 4 + 8 = x^2 - 5x + 12
\]

So, we have:
\[
\frac{x^2 - 5x + 12}{(x-2)(x-4)} = \frac{x+3}{x-4}
\]

Equate the numerators:
\[
x^2 - 5x + 12 = (x+3)(x-2) \implies x^2 - 5x + 12 = x^2 + x - 6 \implies -5x + 12 = x - 6 \implies -6x = -18 \implies x = 3
\]

### Problem 22:
\[
\frac{2}{x+1} + \frac{1}{x-2} = \frac{3x-11}{(x+1)(x-2)}
\]

Find a common denominator:
\[
\frac{2(x-2) + 1(x+1)}{(x+1)(x-2)} = \frac{3x-11}{(x+1)(x-2)}
\]

Simplify the numerator:
\[
2(x-2) + (x+1) = 2x - 4 + x + 1 = 3x - 3
\]

So, we have:
\[
\frac{3x - 3}{(x+1)(x-2)} = \frac{3x-11}{(x+1)(x-2)}
\]

Equate the numerators:
\[
3x - 3 = 3x - 11 \implies -3 = -11 \implies \text{No solution}
\]

### Problem 23:
\[
\frac{x}{x-1} + \frac{x}{x+2} = \frac{x+1}{(x-1)(x+2)}
\]

Find a common denominator:
\[
\frac{x(x+2) + x(x-1)}{(x-1)(x+2)} = \frac{x+1}{(x-1)(x+2)}
\]

Simplify the numerator:
\[
x(x+2) + x(x-1) = x^2 + 2x + x^2 - x = 2x^2 + x
\]

So, we have:
\[
\frac{2x^2 + x}{(x-1)(x+2)} = \frac{x+1}{(x-1)(x+2)}
\]

Equate the numerators:
\[
2x^2 + x = x + 1 \implies 2x^2 = 1 \implies x^2 = \frac{1}{2} \implies x = \pm \frac{1}{\sqrt{2}}
\]

### Problem 24:
\[
\frac{2}{x-4} + \frac{3}{x-3} = \frac{3x-18}{(x-4)(x-3)}
\]

Find a common denominator:
\[
\frac{2(x-3) + 3(x-4)}{(x-4)(x-3)} = \frac{3x-18}{(x-4)(x-3)}
\]

Simplify the numerator:
\[
2(x-3) + 3(x-4) = 2x - 6 + 3x - 12 = 5x - 18
\]

So, we have:
\[
\frac{5x - 18}{(x-4)(x-3)} = \frac{3x-18}{(x-4)(x-3)}
\]

Equate the numerators:
\[
5x - 18 = 3x - 18 \implies 2x = 0 \implies x = 0
\]

### Problem 25:
\[
\frac{x}{x-5} + \frac{3}{x-8} = \frac{-5x+1}{(x-5)(x-8)}
\]

Find a common denominator:
\[
\frac{x(x-8) + 3(x-5)}{(x-5)(x-8)} = \frac{-5x+1}{(x-5)(x-8)}
\]

Simplify the numerator:
\[
x(x-8) + 3(x-5) = x^2 - 8x + 3x - 15 = x^2 - 5x - 15
\]

So, we have:
\[
\frac{x^2 - 5x - 15}{(x-5)(x-8)} = \frac{-5x+1}{(x-5)(x-8)}
\]

Equate the numerators:
\[
x^2 - 5x - 15 = -5x + 1 \implies x^2 - 16 = 1 \implies x^2 = 17 \implies x = \pm \sqrt{17}
\]

### Problem 26:
\[
\frac{x}{x+6} + \frac{3}{x-8} = \frac{-12x+33}{(x+6)(x-8)}
\]

Find a common denominator:
\[
\frac{x(x-8) + 3(x+6)}{(x+6)(x-8)} = \frac{-12x+33}{(x+6)(x-8)}
\]

Simplify the numerator:
\[
x(x-8) + 3(x+6) = x^2 - 8x + 3x + 18 = x^2 - 5x + 18
\]

So, we have:
\[
\frac{x^2 - 5x + 18}{(x+6)(x-8)} = \frac{-12x+33}{(x+6)(x-8)}
\]

Equate the numerators:
\[
x^2 - 5x + 18 = -12x + 33 \implies x^2 + 7x - 15 = 0
\]

Solve the quadratic equation:
\[
x = \frac{-7 \pm \sqrt{49 + 60}}{2} = \frac{-7 \pm \sqrt{109}}{2}
\]

So, the solutions are:
\[
x = \frac{-7 + \sqrt{109}}{2} \quad \text{and} \quad x = \frac{-7 - \sqrt{109}}{2}
\]