ĐKXĐ: \(x\ne0;\pm1\)
\(\left(\dfrac{\left(x+2\right)\left(x+1\right)}{\left(x-1\right)^2}-\dfrac{\left(x-2\right)\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}\right)\dfrac{1}{x}=2\)
\(\Leftrightarrow\dfrac{x^2+3x+2}{\left(x-1\right)^2}-\dfrac{x-2}{x-1}=2x\)
\(\Leftrightarrow\dfrac{x^2-2x+1+5x-5+6}{\left(x-1\right)^2}-\dfrac{x-1-1}{x-1}=2x\)
\(\Leftrightarrow1+\dfrac{5}{x-1}+\dfrac{6}{\left(x-1\right)^2}-1+\dfrac{1}{x-1}=2x\)
\(\Leftrightarrow\dfrac{3}{x-1}+\dfrac{3}{\left(x-1\right)^2}=x=x-1+1\)
Đặt \(\dfrac{1}{x-1}=a\) phương trình trở thành:
\(3a+3a^2=\dfrac{1}{a}+1=\dfrac{a+1}{a}\)
\(\Leftrightarrow3a\left(a+1\right)-\dfrac{a+1}{a}=0\)
\(\Leftrightarrow\left(a+1\right)\left(3a-\dfrac{1}{a}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=-1\\3a=\dfrac{1}{a}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}a=-1\\a=\dfrac{\sqrt{3}}{3}\\a=\dfrac{-\sqrt{3}}{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{x-1}=-1\\\dfrac{1}{x-1}=\dfrac{\sqrt{3}}{3}\\\dfrac{1}{x-1}=\dfrac{-\sqrt{3}}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\left(l\right)\\x=1+\sqrt{3}\\x=1-\sqrt{3}\end{matrix}\right.\)
