Question

Tìm m để PT sau vô nghiệm: \(\frac{1-x}{x-m}+\frac{x-2}{x+m}=\frac{2\left(x-m\right)-2}{m^2-x^2}\)

Answer 1

\(\frac{1-x}{x-m}+\frac{x-2}{x+m}=\frac{2\left(x-m\right)-2}{m^2-x^2}\)(ĐK:\(x\ne\pm m\))

\(\Leftrightarrow\frac{\left(1-x\right)\left(x+m\right)+\left(x-2\right)\left(x-m\right)}{\left(x+m\right)\left(x-m\right)}-\frac{2\left(x-m\right)-2}{m^2-x^2}=0\)

\(\Leftrightarrow\frac{x+m-x^2-mx+x^2-mx-2x+2m}{x^2-m^2}+\frac{2x-2m-2}{x^2-m^2}=0\)

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

\(\Leftrightarrow\frac{-2m.x+m-2}{x^2-m^2}=0\)

\(\Rightarrow-2m.x+m-2=0\)

\(\Leftrightarrow x=\frac{m-2}{2m}\)

Để pt vô nghiệm thì \(\frac{m-2}{2m}\) không xác định

Suy ra:\(2m=0\)

Nên \(m=0\)