Question

Answer 1

\(a,x\left(x-3\right)+2\left(x-3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\\ b,ĐKXĐ:x\ne\pm2\\ \dfrac{x+1}{x-2}+\dfrac{x-3}{x+2}=\dfrac{2\left(x^2-2\right)}{x^2-4}\\ \Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(x-3\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{2x^2-4}{\left(x+2\right)\left(x-2\right)}=0\\ \Leftrightarrow\dfrac{x^2+3x+2+x^2-5x+6-2x^2+4}{\left(x+2\right)\left(x-2\right)}=0\\ \Rightarrow-2x+12=0\\ \Leftrightarrow x=6\left(tm\right)\)