1 câu thôi ạ cần gấp

Đáp án:

ĐK :$\left\{{}\begin{matrix}x+1\ne0\\x\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne-1\\x\ne0\end{matrix}\right.$

 `(x+3)/(x+1) + (x-2)/x =2`

$\Leftrightarrow\dfrac{x\left(x+3\right)+\left(x-2\right)\left(x+1\right)}{x\left(x+1\right)}=\dfrac{2x\left(x+1\right)}{x\left(x+1\right)}$

$\Leftrightarrow x\left(x+3\right)+\left(x-2\right)\left(x+1\right)=2x\left(x+1\right)$

$\Leftrightarrow x^2+3x+\left(x^2+x-2x-2\right)=2x^2+2x$

$\Leftrightarrow x^2+3x+x^2+x-2x-2-2x^2-2x=0$

`<=> (x^2 + x^2 -2x^2)+(3x+x-2x-2x)-2=0`

`<=> -2=0` `(` vô lý `)`

Vậy $S=\varnothing$