Giải thích các bước giải:
a.ĐKXĐ: $x\ne -3$
Ta có:
$x+1+\dfrac{2}{x+3}=\dfrac{x+5}{x+3}$
$\to x+1=\dfrac{x+5}{x+3}-\dfrac{2}{x+3}$
$\to x+1=\dfrac{x+5-2}{x+3}$
$\to x+1=\dfrac{x+3}{x+3}$
$\to x+1=1$
$\to x=0$
b.ĐKXĐ $x\ne 0,1$
Ta có:
$\dfrac{x+3}{x(x-1)}+\dfrac{3}{x}=\dfrac{2-x}{x-1}$
$\to x+3+3\left(x-1\right)=x\left(2-x\right)$
$\to 4x=2x-x^2$
$\to x^2+2x=0$
$\to x(x+2)=0$
$\to x+2=0$ vì $x\ne 0$
$\to x=-2$
