Question

Giải phương trình \(\frac{x+4}{x-3}\) - \(\frac{2}{x}\) = \(\frac{3x+6}{x\left(x-3\right)}\) (ĐKXĐ: x≠0 và x ≠3)

Answer 1

ĐKXĐ: x≠0; x≠3

Ta có: \(\frac{x+4}{x-3}-\frac{2}{x}=\frac{3x+6}{x\left(x-3\right)}\)

\(\Leftrightarrow\frac{x+4}{x-3}-\frac{2}{x}-\frac{3x+6}{x\left(x-3\right)}=0\)

\(\Leftrightarrow\frac{x\left(x+4\right)}{x\left(x-3\right)}-\frac{2\left(x-3\right)}{x\left(x-3\right)}-\frac{3x+6}{x\left(x-3\right)}=0\)

\(\Leftrightarrow x^2+4x-2\left(x-3\right)-\left(3x+6\right)=0\)

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

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

\(\Leftrightarrow x\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=1\end{matrix}\right.\Leftrightarrow x=1\)

Vậy: x=1