Sửa đề: \(9\left(\frac{x-2}{x+1}\right)^2+\left(\frac{x+2}{x-1}\right)^2-10\cdot\frac{x^2-4}{x^2-1}=0\) (1)
ĐKXĐ: x∉{1;-1}
Đặt \(\frac{x-2}{x+1}=a;\frac{x+2}{x-1}=b\)
=>\(a\cdot b=\frac{\left(x-2\right)\left(x+2\right)}{\left(x-1\right)\left(x+1\right)}=\frac{x^2-4}{x^2-1}\)
(1) sẽ trở thành: \(9a^2+b^2-10ab=0\)
=>(9a-b)(a-b)=0
TH1: 9a-b=0
=>b=9a
=>\(\frac{x+2}{x-1}=9\cdot\frac{x-2}{x+1}=\frac{9x-18}{x+1}\)
=>\(\left(x+2\right)\left(x+1\right)=\left(9x-18\right)\left(x-1\right)\)
=>\(9x^2-9x-18x+18=x^2+3x+2\)
=>\(8x^2-30x+16=0\)
=>\(4x^2-15x+8=0\)
=>\(x^2-\frac{15}{4}x+2=0\)
=>\(x^2-2\cdot x\cdot\frac{15}{8}+\frac{225}{64}=\frac{97}{64}\)
=>\(\left(x-\frac{15}{8}\right)^2=\frac{97}{64}\)
=>\(\left[\begin{array}{l}x-\frac{15}{8}=\frac{\sqrt{97}}{8}\\ x-\frac{15}{8}=-\frac{\sqrt{97}}{8}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\sqrt{97}+15}{8}\left(nhận\right)\\ x=\frac{-\sqrt{97}+15}{8}\left(nhận\right)\end{array}\right.\)
TH2: a-b=0
=>a=b
=>\(\frac{x-2}{x+1}=\frac{x+2}{x-1}\)
=>(x-2)(x-1)=(x+2)(x+1)
=>\(x^2-3x+2=x^2+3x+2\)
=>-6x=0
=>x=0(nhận)
