ta có: \(\frac{1}{x^2-2x+2}+\frac{2}{x^2-2x+3}=\frac{6}{x^2-2x+4}\)(1)
đặt \(x^2-2x+3=a\)(a\(\ge2\))
khi đó:
(1)\(\Leftrightarrow\frac{1}{a-1}+\frac{2}{a}=\frac{6}{a+1}\)
\(\Leftrightarrow\frac{a\left(a+1\right)}{a\left(a-1\right)\left(a+1\right)}+\frac{2\left(a-1\right)\left(a+1\right)}{a\left(a-1\right)\left(a+1\right)}-\frac{6a\left(a-1\right)}{a\left(a-1\right)\left(a+1\right)}=0\)
\(\Leftrightarrow\frac{a^2+a+2a^2-2-6a^2+6a}{a\left(a-1\right)\left(a+1\right)}=0\)
\(\Leftrightarrow-3a^2+7a-2=0\) (vì \(a\left(a-1\right)\left(a+1\right)\ne0\))
\(\Leftrightarrow\left(1-3a\right)\left(a-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}1-3a=0\\a-2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}a=\frac{1}{3}\left(loại\right)\\a=2\end{matrix}\right.\)
\(\Leftrightarrow x^2-2x+3=2\Leftrightarrow\left(x-1\right)^2=0\)
\(\Leftrightarrow x=1\)
vậy x=1 là nghiệm của phương trình
