`\frac{x-1}{x+2}-\frac{x}{x-2}=\frac{5x-8}{x^2-4}`

Đáp án+Giải thích các bước giải:

$\dfrac{x-1}{x+2}-$$\dfrac{x}{x-2}=$$\dfrac{5x-8}{x^2-4}$           `9ĐKXĐ: x\ne±2)`

`<=>` $\dfrac{x-1}{x+2}-$$\dfrac{x}{x-2}=$$\dfrac{5x-8}{(x+2)(x-2)}$ 

`<=>` $\dfrac{(x-1)(x-2)}{(x+2)(x-2)}-$$\dfrac{x(x+2)}{(x-2)(x+2)}=$$\dfrac{5x-8}{(x+2)(x-2)}$ 

`<=>` $(x-1)(x-2)-x(x+2)=5x-8$

`<=>` $x^2-2x-x+2-x^2-2x=5x-8$

`<=>` $-5x+2=5x-8$

`<=>` $-5x-5x=-8-2$

`<=>` $-10x=-10$

`<=>` $x=1(t.m)$

Vậy `S={1}`