\(\frac{4}{x^2-3x+2}-\frac{3}{2x^2-6x+1}+1=0\) \(Đkxđ:.......\)
Đặt: \(t=x^2-3x+2\left(t\ne0\right)\)
\(\Rightarrow2t=2x^2-6x+4\)
\(\Rightarrow2x^2-6x+1=2t-3\)
\(Pt:\Leftrightarrow\frac{4}{7}-\frac{3}{2t-3}+1=0\)
\(\Leftrightarrow4\left(2t-3\right)-3t+t\left(2t-3\right)=0\)
\(\Leftrightarrow8t-12-3t+2t^2-3t=0\)
\(\Leftrightarrow2t^2+2t-12=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=2\\t=-3\end{matrix}\right.\left(tm:\left[{}\begin{matrix}t\ne0\\t\ne\frac{3}{2}\end{matrix}\right.\right)\)
+ Với \(t=2\) thì: \(x^2-3x+2=2\)
\(\Leftrightarrow x\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\left(tmđk\right)\)
+ Với \(t=-3\) thì \(x^2-3x+2=-3\)
\(\Leftrightarrow x^2-2.\frac{3}{2}x+\frac{9}{4}+\frac{11}{4}=0\)
\(\Leftrightarrow\left(x-\frac{3}{2}\right)^2+\frac{11}{4}=0\left(vô-lí\right)\)
Vậy pt có nghiệm: \(\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
Bài 2:
ĐKXĐ: $x\neq 1;2;3;6$
PT $\Leftrightarrow \frac{2}{x-2}+\frac{3}{x-3}=\frac{6}{x-6}-\frac{1}{x-1}$
$\Leftrightarrow \frac{5x-12}{x^2-5x+6}=\frac{5x}{x^2-7x+6}$
Đặt $x^2+6=t$ thì $\frac{5x-12}{t-5x}=\frac{5x}{t-7x}$
$\Rightarrow (5x-12)(t-7x)=5x(t-5x)$
$\Leftrightarrow 10x^2+12t+84x=0$
$\Leftrightarrow 10x^2+12(x^2+6)+84x=0$
$\Leftrightarrow 22x^2+84x+72=0$
$\Leftrightarrow 11x^2+42x+36=0$
$\Rightarrow x=\frac{-21\pm 3\sqrt{5}}{11}$
