Giải thích các bước giải:
a.ĐKXĐ:$ x\ne 0, 1$
Ta có:
$\dfrac{2}{x-1}+\dfrac{3(x-1)}{x}=5$
$\to 2x+3\left(x-1\right)^2=5x\left(x-1\right)$
$\to 3x^2-4x+3=5x^2-5x$
$\to 2x^2-x-3=0$
$\to (2x-3)(x+1)=0$
$\to x\in\{\dfrac32,-1\}$
b.ĐKXĐ: $x\ne 0, -1$
Ta có:
$\dfrac{x-1}{x}+\dfrac{1}{x+1}=\dfrac{2x-1}{x^2+x}$
$\to\dfrac{x-1}{x}+\dfrac{1}{x+1}=\dfrac{2x-1}{x(x+1)}$
$\to (x-1)(x+1)+x=2x-1$
$\to x^2+x-1=2x-1$
$\to x^2-x=0$
$\to x(x-1)=0$
$\to x=1$ vì $x\ne 0$
c.ĐKXĐ: $x\ne \pm1$
Ta có:
$\dfrac{x-1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{16}{x^2-1}$
$\to 1-\dfrac{x-1}{x+1}=\dfrac{16}{(x-1)(x+1)}$
$\to (x-1)(x+1)-(x-1)^2=16$
$\to 2x-2=16$
$\to x=9$
d.ĐKXĐ: $x\ne -2, 3$
Ta có:
$1+\dfrac{x}{3-x}=\dfrac{5x}{(x+2)(3-x)}+\dfrac{2}{x+2}$
$\to \left(3-x\right)\left(x+2\right)+x\left(x+2\right)=5x+2\left(3-x\right)$
$\to x-x^2+6+x^2+2x+x^2+2x=5x+6-2x$
$\to 3x+6=3x+6$ luôn đúng
$\to$Phương trình có vô số nghiệm $x\ne 3, -2$
