Question

Giải PT:

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

Answer 1

ĐKXĐ: \(x\ne\left\{3;4;5;6\right\}\)

\(\Leftrightarrow\frac{x\left(x-6\right)+x\left(x-3\right)}{\left(x-3\right)\left(x-6\right)}=\frac{x\left(x-4\right)+x\left(x-5\right)}{\left(x-4\right)\left(x-5\right)}\)

\(\Leftrightarrow\frac{2x^2-9x}{\left(x-3\right)\left(x-6\right)}=\frac{2x^2-9x}{\left(x-4\right)\left(x-5\right)}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x^2-9x=0\\\left(x-3\right)\left(x-6\right)=\left(x-4\right)\left(x-5\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\left(2x-9\right)=0\\18=20\left(vn\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{9}{2}\end{matrix}\right.\)