Ta có: \(\frac{1}{x-1}+\frac{1}{x-2}=\frac{1}{x+2}+\frac{1}{x+1}\)
\(\Leftrightarrow\frac{x-2}{\left(x-1\right)\left(x-2\right)}+\frac{x-1}{\left(x-2\right)\left(x-1\right)}=\frac{x+1}{\left(x+2\right)\left(x+1\right)}+\frac{x+2}{\left(x+1\right)\left(x+2\right)}\)
\(\Leftrightarrow\frac{2x-3}{\left(x-1\right)\left(x-2\right)}-\frac{2x+3}{\left(x+1\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{\left(2x-3\right)\left(x+1\right)\left(x+2\right)}{\left(x-1\right)\left(x-2\right)\left(x+1\right)\left(x+2\right)}-\frac{\left(2x+3\right)\left(x-1\right)\left(x-2\right)}{\left(x-1\right)\left(x-2\right)\left(x+1\right)\left(x+2\right)}=0\)
\(\Leftrightarrow2x^3+3x^2+10x-6-\left(2x^3-3x^2-5x+6\right)=0\)
\(\Leftrightarrow2x^3+3x^2+10x-6-2x^3+3x^2+5x-6=0\)
\(\Leftrightarrow6x^2+15x-12=0\)
\(\Leftrightarrow6\left(x^2+\frac{5}{2}x-2\right)=0\)
Vì 6>0
nên \(x^2+\frac{5}{2}x-2=0\)
\(\Leftrightarrow x^2+2\cdot x\cdot\frac{5}{4}+\frac{25}{16}-\frac{57}{16}=0\)
\(\Leftrightarrow\left(x+\frac{5}{4}\right)^2-\frac{57}{16}=0\)
\(\Leftrightarrow\left(x+\frac{5}{4}\right)^2=\frac{57}{16}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{5}{4}=\sqrt{\frac{57}{16}}\\x+\frac{5}{4}=-\sqrt{\frac{57}{16}}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-5+\sqrt{57}}{4}\\x=-\frac{5+\sqrt{57}}{4}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{-5+\sqrt{57}}{4};-\frac{5+\sqrt{57}}{4}\right\}\)
