Giải thích các bước giải:
a.ĐKXĐ $x\ne\pm1$
Ta có:
$\dfrac{2x+1}{x-1}=\dfrac{5(x-1)}{x+1}$
$\to (2x+1)(x+1)=5(x-1)^2$
$\to 2x^2+3x+1=5x^2-10x+5$
$\to 3x^2-13x+4=0$
$\to (3x-1)(x-4)=0$
$\to x\in\{\dfrac13,4\}$
b.ĐKXĐ: $x\ne 2, 4$
Ta có:
$\dfrac{x-3}{x-2}+\dfrac{x-2}{x-4}=-1$
$\to \left(x-3\right)\left(x-4\right)+\left(x-2\right)^2=-\left(x-2\right)\left(x-4\right)$
$\to 2x^2-11x+16=-x^2+6x-8$
$\to 3x^2-17x+24=0$
$\to (x-3)(3x-8)=0$
$\to x\in\{3, \dfrac83\}$
