\(\dfrac{1}{x-1}-\dfrac{2}{2-x}=\dfrac{5}{\left(x-1\right)\left(x-2\right)}\)
\(\Leftrightarrow\dfrac{1}{x-1}+\dfrac{2}{x-2}=\dfrac{5}{\left(x-1\right)\left(x-2\right)}\)
ĐKXĐ : \(\left\{{}\begin{matrix}x-1\ne0\\x-2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ne2\end{matrix}\right.\)
Ta có : \(\dfrac{1}{x-1}+\dfrac{2}{x-2}=\dfrac{5}{\left(x-1\right)\left(x-2\right)}\)
\(\Leftrightarrow\dfrac{x-2}{\left(x-1\right)\left(x-2\right)}+\dfrac{2\left(x-1\right)}{\left(x-2\right)\left(x-1\right)}=\dfrac{5}{\left(x-1\right)\left(x-2\right)}\)
`=> x-2+2(x-1)=5`
`<=> x-2+2x-2=5`
`<=> 3x-4=5`
`<=> 3x=9`
`<=>x=3` ( thỏa mãn đk )
Vậy pt đã cho có nghiệm `x=3`
` @` Đề như này nhỉ ^^
\(chucbanhoctot\)
