\(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\) \(\left(ĐKXĐ:x\ne-1;x\ne2\right)\)
\(\Leftrightarrow\frac{x-2}{\left(x+1\right)\left(x-2\right)}-\frac{5x+5}{\left(x+1\right)\left(x-2\right)}=\frac{-15}{\left(x+1\right)\left(x-2\right)}\)
\(\Leftrightarrow-4x-7=-15\Leftrightarrow4x=8\Leftrightarrow x=2\left(loại\right)\)
Vậy phương trình vô nghiệm.
ĐKXĐ : \(\left\{{}\begin{matrix}x+1\ne0\\x-2\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ne-1\\x\ne2\end{matrix}\right.\)
- Ta có : \(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)
=> \(\frac{x-2}{\left(x+1\right)\left(x-2\right)}-\frac{5\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}=\frac{-15}{\left(x+1\right)\left(x-2\right)}\)
=> \(x-2-5\left(x+1\right)=-15\)
=> \(x-2-5x-5+15=0\)
=> \(-4x=-8\)
=> \(x=2\) ( KTM )
Vậy phương trình vô nghiệm .
