\(\frac{x-2}{x+2}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{x^2-4}\left(x\ne\pm2\right)\)
\(\Leftrightarrow\frac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}-\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{2x-22}{\left(x-2\right)\left(x+2\right)}\)
=> x2 - 4x + 4 - 3x - 6 = 2x - 22
<=> x2 - 9x + 20 = 0
<=> x2 - 4x - 5x + 20 = 0
<=> x( x - 4 ) - 5( x - 4 ) = 0
<=> ( x - 4 )( x - 5 ) = 0
<=> x - 4 = 0 hoặc x - 5 = 0
<=> x = 4 (tm) hoặc x = 5 (tm)
Vậy ...
\(\frac{x-2}{x+2}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{x^2-4}\)ĐKXĐ : \(x\ne\pm2\)
\(\Leftrightarrow\frac{\left(x-2\right)^2-3\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{2x-22}{\left(x-2\right)\left(x+2\right)}\)
\(\Rightarrow x^2-4x+4-3x-6=2x-22\)
\(\Leftrightarrow x^2-7x-2=2x-22\Leftrightarrow x^2-9x+20=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\Leftrightarrow x=4;x=5\)( tmđk )
Vậy tập nghiệm phương trình là S = { 4 ; 5 }