\(\dfrac{x+2}{x-2}+\dfrac{1}{x}=-\dfrac{8}{2x-x^2}\)
\(\Leftrightarrow\dfrac{x+2}{x-2}+\dfrac{1}{x}=-\dfrac{8}{x\left(2-x\right)}\)
\(\Leftrightarrow\dfrac{x+2}{x-2}+\dfrac{1}{x}=\dfrac{8}{x\left(x-2\right)}\)
ĐKXĐ : \(\left\{{}\begin{matrix}x\ne0\\x-2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne2\end{matrix}\right.\)
ta có : \(\dfrac{x+2}{x-2}+\dfrac{1}{x}=\dfrac{8}{x\left(x-2\right)}\)
\(\Leftrightarrow\dfrac{\left(x+2\right)x}{x\left(x-2\right)}+\dfrac{x-2}{x\left(x-2\right)}=\dfrac{8}{x\left(x-2\right)}\)
`=> (x+2)x +x-2=8`
`<=>x^2+2x+x-2=8`
`<=> x^2 +3x -2-8=0`
`<=> x^2 + 3x -10=0`
`<=>x^2+5x -2x-10=0`
`<=>x(x+5)-2(x+5)=0`
`<=>(x+5)(x-2)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(loai\right)\\x=-5\end{matrix}\right.\)
vậy pt có nghiệm `x=-5`
\(\dfrac{x+2}{x-2}+\dfrac{1}{x}=-\dfrac{8}{2x-x^2}\left(x\ne0;x=\ne2\right)\\ < =>\dfrac{x+2}{x-2}+\dfrac{1}{x}+\dfrac{8}{x\left(2-x\right)}=0\\ < =>\dfrac{x+2}{x-2}+\dfrac{1}{x}-\dfrac{8}{x\left(x-2\right)}=0\)
suy ra
`(x+2)x+x-2=8`
`<=> x^2 +2x+x-2-8=0`
`<=> x^2 + 3x-10=0`
`<=> x^2 +5x-2x-10=0`
`<=> x(x+5)-2(x+5)=0`
`<=> (x+5)(x-2)=0`
\(< =>\left[{}\begin{matrix}x+5=0\\x-2=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=-5\left(tm\right)\\x=2\left(ktmdk\right)\end{matrix}\right.\)
