\(20\left(\frac{x-2}{x+1}\right)^2-5\left(\frac{x+2}{x-1}\right)+48\left(\frac{x^2-4}{x^2-1}\right)\)
ĐK: \(\left|x\right|\ne1\) với \(\left|x\right|=2\Rightarrow V0.N_o\Rightarrow\left|x\right|\ne2\)
Đặt \(\left(\frac{x-2}{x+1}\right)=t\Rightarrow t\ne0\)
\(\Leftrightarrow20t^2-5\left(\frac{1}{t^2}\right)+48=0\Leftrightarrow\left\{\begin{matrix}t^2=y\\y>0\\20y^2+48y-5=0\left(2\right)\end{matrix}\right.\) (I)
\(\left(2\right)\Leftrightarrow y^2+2.\frac{6}{5}y+\left(\frac{6}{5}\right)^{^2}=\left(\frac{13}{10}\right)^2\)
\(\Rightarrow\left\{\begin{matrix}y=\frac{-12-13}{10}=\frac{-5}{2}\left(loai\right)\\y=\frac{-12+13}{10}=\frac{1}{10}\end{matrix}\right.\)
\(\left(I\right)\Rightarrow\left[\begin{matrix}t=-\frac{1}{\sqrt{10}}\\t=\frac{1}{\sqrt{10}}\end{matrix}\right.\) giải tiếp ok!
