Giải thích các bước giải:

a.ĐKXĐ: $x\ne 1, x\ne 2$

$\dfrac4{x-1}-\dfrac5{x-2}=-3$ 

$\to 4\left(x-2\right)-5\left(x-1\right)=-3\left(x-1\right)\left(x-2\right)$

$\to 4\left(x-2\right)-5\left(x-1\right)=-3\left(x-1\right)\left(x-2\right)$

$\to 3x^2-10x+3=0$

$\to (3x-1)(x-3)=0$

$\to x\in\{\dfrac13, 3\}$

b.ĐKXĐ: $x\ne\pm\dfrac13$

$\dfrac{12}{1-9x^2}=\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}$

$\to \dfrac{12}{(1-3x)(1+3x)}=\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}$

$\to 12=(1-3x)^2-(1+3x)^2$

$\to 12=-12x$

$\to x=-1$

c.ĐKXĐ: $x\ne 2, x\ne5$

$\dfrac{6x+1}{x^2-7x+10}+\dfrac5{x-2}=\dfrac3{x-5}$

$\to \dfrac{6x+1}{(x-2)(x-5)}+\dfrac5{x-2}=\dfrac3{x-5}$

$\to 6x+1+5(x-5)=3(x-2)$

$\to 11x-24=3x-6$

$\to 8x=18$

$\to x=\dfrac94$

d.ĐKXĐ: $x\notin\{0,5, -5\}$

Ta có:

$\dfrac{x+5}{x^2-5x}-\dfrac{x+25}{2x^2-50}=\dfrac{x-5}{2x^2+10x}$

$\to \dfrac{x+5}{x(x-5)}-\dfrac{x+25}{2(x-5)(x+5)}=\dfrac{x-5}{2x(x+5)}$

$\to 2(x+5)^2-x(x+25)=(x-5)^2$

$\to x^2-5x+50=x^2-10x+25$

$\to 5x=-25$

$\to x=-5$ loại vì $x\ne -5$