Giải thích các bước giải:
f.ĐKXĐ: $x\ne 2, 3$
Ta có:
$\dfrac{2x}{x-2}+\dfrac{5}{3-x}=\dfrac{5}{x^2-5x+6}$
$\to \dfrac{2x}{x-2}-\dfrac{5}{x-3}=\dfrac{5}{(x-2)(x-3)}$
$\to 2x\left(x-3\right)-5\left(x-2\right)=5$
$\to 2x^2-11x+10=5$
$\to 2x^2-11x+5=0$
$\to (x-5)(2x-1)=0$
$\to x\in\{5, \dfrac12\}$
i.ĐKXD: $x\ne 2, -4$
Ta có:
$\dfrac{x}{x+4}-\dfrac{2x}{x-2}=\dfrac{8(x+1)}{(2-x)(4+x)}$
$\to \dfrac{x}{x+4}-\dfrac{2x}{x-2}=-\dfrac{8(x+1)}{(x-2)(4+x)}$
$\to x\left(x-2\right)-2x\left(x+4\right)=-8\left(x+1\right)$
$\to -x^2-10x=-8x-8$
$\to x^2+2x-8=0$
$\to (x+4)(x-2)=0$
$\to x\in\{-4, 2\}$ mà $x\ne 2, -4$
$\to$Phương trình vô nghiệm
l.Ta có:
$x^2(x^2-10)=-9$
$\to x^2(x^2-9-1)=-9$
$\to x^2(x^2-9)-x^2=-9$
$\to x^2(x^2-9)-x^2+9=0$
$\to x^2(x^2-9)-(x^2-9)=0$
$\to (x^2-1)(x^2-9)=0$
$\to x^2\in\{1, 9\}$
$\to x\in\{1 , -1 , 3, -3\}$
