ĐKXĐ: \(x\ne1;\frac{3}{2}\)
Nhận thấy \(x=0\) không phải nghiệm, pt tương đương:
\(\frac{2}{2x-5+\frac{3}{x}}+\frac{9}{2x-1+\frac{3}{x}}=6\)
Đặt \(2x-5+\frac{1}{x}=a\Rightarrow2x-1+\frac{3}{x}=a+4\) ta được:
\(\frac{2}{a}+\frac{9}{a+4}=6\)
\(\Leftrightarrow2\left(a+4\right)+9a=6a\left(a+4\right)\)
\(\Leftrightarrow6a^2+13a-8=0\Rightarrow\left[{}\begin{matrix}a=\frac{1}{a}\\a=-\frac{8}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x-5+\frac{1}{x}=\frac{1}{2}\\2x-5+\frac{1}{x}=-\frac{8}{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}2x^2-\frac{11}{2}x+1=0\\2x^2-\frac{7}{3}x+1=0\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow x=\frac{11\pm\sqrt{89}}{8}\)
