Question

Giải phương trình: \(\dfrac{2x-4}{x-3}+\dfrac{6}{3x-x^2}=\dfrac{x-1}{x}\)

Answer 1

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

\(\Leftrightarrow\dfrac{2x-4}{x-3}+\dfrac{-6}{x\left(x-3\right)}=\dfrac{x-1}{x}\)

ĐKXĐ: x ≠ 0 và x ≠ 3

\(\Leftrightarrow\dfrac{x\left(2x-4\right)}{x\left(x-3\right)}+\dfrac{-6}{x\left(x-3\right)}=\dfrac{\left(x-3\right)\left(x-1\right)}{x\left(x-3\right)}\)

\(\Leftrightarrow x\left(2x-4\right)+\left(-6\right)=\left(x-3\right)\left(x-1\right)\)

\(\Leftrightarrow2x^2-4x-6=x^2-3x-x+3\)

\(\Leftrightarrow2x^2-4x-6-x^2+3x+x-3=0\)

\(\Leftrightarrow x^2-9=0\Leftrightarrow x^2-3^2=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)