Giải thích các bước giải:
1.Ta có :
$\dfrac{15x-10}{x^2+3}=0$
$\to 15x-10=0$
$\to x=\dfrac{10}{15}=\dfrac 23$
2.ĐKXĐ : $x\ne -1,1$
$\dfrac{x-1}{x+1}-\dfrac{x^2+x-2}{x-1}=\dfrac{x+1}{x-1}-x-2$
$\to \dfrac{x-1}{x+1}-\dfrac{(x+2)(x-1)}{x-1}=\dfrac{x+1}{x-1}-x-2$
$\to \dfrac{x-1}{x+1}-(x+2)=\dfrac{x+1}{x-1}-(x+2)$
$\to \dfrac{x-1}{x+1}=\dfrac{x+1}{x-1}$
$\to (x-1)^2=(x+1)^2$
$\to x^2-2x+1=x^2+2x+1$
$\to 4x=0$
$\to x=0$