Đáp án:j)$x=\frac{2}{5}$
l)$ 4(x^{2}+x-1)=0$
Giải thích các bước giải:
$j)\frac{x-1}{x+2}-\frac{x}{x-2}=\frac{5x-2}{4-x^{2}} (đk: x\neq2, x\neq-2)$
$<=>\frac{x-1}{x+2}-\frac{x}{x-2}=-\frac{5x-2}{x^{2}-4}$
$=> (x-1)(x-2)-x(x+2)=5x-2$
$<=> x^{2}-3x+2-x^{2}-2x=5x-2$
$<=> 10x=4 <=> x=\frac{2}{5}$
$l)\frac{x-1}{x+1}-\frac{x^{2}+x-2}{x+1}=\frac{x+1}{x-1}-x-2 (đk: x\neq1, x\neq-1)$
$<=> \frac{x-1}{x+1}-\frac{(x-1)(x+2)}{x+1}=\frac{x+1}{x-1}-x-2$
$=> (x-1)(x-1)-(x-1)(x^{2}+x-2)=(x+1)(x+1)-(x-2)(x^{2}-1)$
$<=> x^{2}-2x+1-(x^{3}+2x^{2}-x-2)=x^{2}+2x+1-(x^{3}-2x^{2}-x+2)$
$<=> x^{2}-2x+1-x^{3}-2x^{2}+x+2=x^{2}+2x+1-x^{3}+2x^{2}+x-2$
$<=> 4x^{2}+4x-4=0$
$<=> 4(x^{2}+x-1)=0$
