bBài 52a giúp mk vs ạ
ĐKXĐ: \(x\ne0;1;2\)
\(\Leftrightarrow\frac{x\left(4x-7\right)}{x\left(x^2-3x+2\right)}=\frac{9x^2-16x+4}{x\left(x^2-3x+2\right)}\)
\(\Leftrightarrow4x^2-7x=9x^2-16x+4\)
\(\Leftrightarrow5x^2-9x+4=0\)
\(\Leftrightarrow\left(5x-4\right)\left(x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{4}{5}\\x=1\left(l\right)\end{matrix}\right.\)