Question

Giải phương trình sau

\(\frac{1}{x^2-3x+3}+\frac{2}{x^2-3x+4}=\frac{6}{x^2-3x+5}\)

Answer 1

Đặt \(x^2-3x+4=a\) phương trình trở thành:

\(\frac{1}{a-1}+\frac{2}{a}=\frac{6}{a+1}\Leftrightarrow\frac{6}{a+1}-\frac{1}{a-1}-\frac{2}{a}=0\)

\(\Leftrightarrow\frac{5a-7}{a^2-1}-\frac{2}{a}=0\Leftrightarrow\frac{5a^2-7a-2a^2+2}{a\left(a^2-1\right)}=0\)

\(\Leftrightarrow3a^2-7a+2=0\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{1}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2-3x+4=2\\x^2-3x+4=\frac{1}{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2-3x+2=0\\x^2-3x+\frac{11}{3}=0\left(vn\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)