giúp mình với ạ
\(\dfrac{x+10}{x-2}-\dfrac{4}{x}=-\dfrac{8}{x\left(2-x\right)}\)
`<=>`\(\dfrac{x+10}{x-2}-\dfrac{4}{x}=\dfrac{8}{x\left(x-2\right)}\)
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 :\(\dfrac{x+10}{x-2}-\dfrac{4}{x}=\dfrac{8}{x\left(x-2\right)}\)
\(\Leftrightarrow\dfrac{x\left(x+10\right)}{x\left(x-2\right)}-\dfrac{4\left(x-2\right)}{x\left(x-2\right)}=\dfrac{8}{x\left(x-2\right)}\)
`=> x^2 +10x -4x+8 =8`
`<=> x^2 + 6x=0`
`<=> x(x+6)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(l\right)\\x=-6\left(tm\right)\end{matrix}\right.\)
Vậy pt có nghiệm `x=-6`
=>x(x+10)-4(x-2)=8
=>x^2+10x-4x+8=8
=>x^2+6x=0
=>x=0(loại) hoặc x=-6(nhận)
giúp mình với ạ
