Question

[Dung Vu]

Giải phương trình sau : \(\dfrac{x^2+3x+2}{x-3}\left(\dfrac{x+1}{x-2}+1\right)=\dfrac{x^2+3x+2}{x-3}.\dfrac{x^2}{2-x}\)

Answer 1

[nthv_.]

\(ĐK:x\ne3;x\ne2\\ PT\Leftrightarrow\dfrac{x^2+3x+2}{x-3}\left(\dfrac{x+1}{x-2}+1+\dfrac{x^2}{x-2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\dfrac{\left(x+1\right)\left(x+2\right)}{x-3}=0\\\dfrac{x^2+x+2}{x-2}=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\\x^2+x+2=0\left(vô.n_0\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\end{matrix}\right.\)