\(\frac{x-2}{x+2}+\frac{3}{2-x}=\frac{2\left(x-11\right)}{x^2-4}\) (Đkxd: tự tìm)
\(\Leftrightarrow\frac{\left(x-2\right)^2\left(2-x\right)}{\left(x+2\right)\left(x-2\right)\left(2-x\right)}+\frac{3\left(x+2\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)\left(2-x\right)}=\frac{2\left(x-11\right)\left(2-x\right)}{\left(x-2\right)\left(x+2\right)\left(2-x\right)}\)
\(\Leftrightarrow9x^2-x^3-12x-4-26x+2x^2+44=0\)
\(\Leftrightarrow11x^2-x^3-38x+40=0\)
\(\Leftrightarrow\left(-x^2+9x-20\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x^2-9x+20\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-4\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ne5\left(tm\right)\\x\ne4\left(tm\right)\\x\ne2\left(ktm\right)\end{matrix}\right.\)
Vậy ........
