Question

[Thảo Vy]

\(M=\frac{\left(x-3\right)^2}{2x^2-6x}\left(1-\frac{6x+18}{x^2-9}\right)\)

Rút gọn biểu thức

Answer 1

[ღ๖ۣۜLinh's ๖ۣۜLinh'sღ] ★we are one★]

ĐKXĐ \(\hept{\begin{cases}x\ne0\\x\ne\pm3\end{cases}}\)

\(M=\frac{\left(x-3\right)^2}{2x\left(x-3\right)}\left(1-\frac{6\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\right)\)

\(=\frac{x-3}{2x}\left(1-\frac{6}{x-3}\right)\)

\(=\frac{x-3}{2x}.\frac{x-9}{x-3}=\frac{x-9}{2x}\)

Answer 2

[Tran Le Khanh Linh]

\(M=\frac{\left(x-3\right)^2}{2x^2-6x}\left(1-\frac{6x+18}{x^2-9}\right)\left(x\ne\pm3;x\ne0\right)\)

\(\Leftrightarrow M=\frac{\left(x-3\right)^2}{2x\left(x-3\right)}\left(1-\frac{6\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\right)\)

\(\Leftrightarrow M=\frac{x-3}{2x}\cdot\left(1-\frac{6}{x-3}\right)\)

\(\Leftrightarrow M=\frac{x-3}{2x}\cdot\frac{x-9}{x-3}\)

\(\Leftrightarrow M=\frac{x-9}{2x}\)

Vậy với \(x\ne\pm3;x\ne0\)thì \(M=\frac{x-9}{2x}\)