`9/[x^2-4]=[x-1]/[x+2]+3/[x-2]` `ĐK: x \ne +-2`
`<=>9/[(x-2)(x+2)]=[(x-1)(x-2)+3(x+2)]/[(x-2)(x+2)]`
`=>9=x^2-2x-x+2+3x+6`
`<=>x^2=1`
`<=>x=+-1` (t/m)
Vậy `x=+-1`
\(\dfrac{9}{x^2-4}=\dfrac{x-1}{x+2}+\dfrac{3}{x-2}\left(đkxđ:x\ne\pm2\right)\\ \Leftrightarrow\dfrac{9}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x-1\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\\ \Rightarrow9=x^2-3x+2+3x+6\\ \Leftrightarrow x^2=1\\ \Leftrightarrow x^2=\pm1\left(TM\right)\)
Vậy PT có tập nghiệm \(S=\left\{-1;1\right\}\)
\(\Leftrightarrow x^2-3x+2+3x+6=9\)
\(\Leftrightarrow x^2=1\)
=>x=1 hoặc x=-1
ĐKXĐ: \(x\ne\pm2\).
\(\dfrac{9}{x^2-4}=\dfrac{x-1}{x+2}+\dfrac{3}{x-2}\)
\(\Leftrightarrow\dfrac{9}{\left(x-2\right)\left(x+2\right)}=\dfrac{x-1}{x+2}+\dfrac{3}{x-2}\)
\(\Leftrightarrow\dfrac{9}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x-1\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(\Rightarrow x^2-2x-x+2+3x+6=9\)
\(\Leftrightarrow x^2+2+6-9=0\)
\(\Leftrightarrow x^2-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-1\left(nhận\right)\end{matrix}\right.\)
-Vậy \(S=\left\{\pm1\right\}\)
