Giải thích các bước giải:

Bài 17:

$R=\dfrac{5x}{2x^2-3x}+\dfrac2{2x+3}-\dfrac{2x-33}{9-4x^2}$

$\to R=\dfrac5{2x-3}+\dfrac2{2x+3}+\dfrac{2x-33}{4x^2-9}$ 

$\to R=\dfrac{5\left(2x+3\right)}{\left(2x-3\right)\left(2x+3\right)}+\dfrac{2\left(2x-3\right)}{\left(2x-3\right)\left(2x+3\right)}+\dfrac{2x-33}{\left(2x-3\right)\left(2x+3\right)}$

$\to R=\dfrac{5\left(2x+3\right)+2\left(2x-3\right)+2x-33}{\left(2x-3\right)\left(2x+3\right)}$

$\to R=\dfrac{8}{2x+3}$

Bài 18:

$S=2y-\dfrac{6xy+2y}{3x+2y}+\dfrac{2y-9x^2}{3x+2y}$

$\to S=\dfrac{2y(3x+2y)-(6xy+2y)+2y-9x^2}{3x+2y}$

$\to S=\dfrac{4y^2-9x^2}{3x+2y}$

$\to S=\dfrac{(2y-3x)(2y+3x)}{3x+2y}$

$\to S=2y-3x$

Bài 19:

$T=\dfrac4{x+2}+\dfrac3{x-2}+\dfrac{5x+2}{4-x^2}-\dfrac{x^2-2x+4}{x^3+8}$

$\to T=\dfrac4{x+2}+\dfrac3{x-2}-\dfrac{5x+2}{x^2-4}-\dfrac{x^2-2x+4}{(x+2)(x^2-2x+4)}$

$\to T=\dfrac4{x+2}+\dfrac3{x-2}-\dfrac{5x+2}{x^2-4}-\dfrac{1}{x+2}$

$\to T=\dfrac3{x+2}+\dfrac3{x-2}-\dfrac{5x+2}{x^2-4}$

$\to T=\dfrac{3\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\dfrac{3\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{5x+2}{\left(x+2\right)\left(x-2\right)}$

$\to T=\dfrac{3\left(x-2\right)+3\left(x+2\right)-\left(5x+2\right)}{\left(x+2\right)\left(x-2\right)}$

$\to T=\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}$

$\to T=\dfrac{1}{x+2}$

20.Ta có:

$U=\dfrac{1-2x}{2x}+\dfrac{2x}{2x-1}+\dfrac1{2x-4x^2}$

$\to U=\dfrac{1-2x}{2x}+\dfrac{2x}{2x-1}-\dfrac1{4x^2-2x}$

$\to U=\dfrac{\left(1-2x\right)\left(2x-1\right)}{2x\left(2x-1\right)}+\dfrac{4x^2}{2x\left(2x-1\right)}-\dfrac{1}{2x\left(2x-1\right)}$

$\to U=\dfrac{\left(1-2x\right)\left(2x-1\right)+4x^2-1}{2x\left(2x-1\right)}$

$\to U=\dfrac{4x-2}{2x\left(2x-1\right)}$

$\to U=\dfrac1x$