1/\(m=1\) pt vô nghiệm (ktm)
Với \(m\ne1\Rightarrow\left(m-1\right)x=-3m+2\Rightarrow x=\frac{-3m+2}{m-1}\)
\(\Rightarrow\frac{-3m+2}{m-1}\ge1\Leftrightarrow\frac{-3m+2}{m-1}-1\ge0\Leftrightarrow\frac{-4m+3}{m-1}\ge0\)
\(\Rightarrow\frac{3}{4}\le m< 1\)
Câu 2:
ĐKXĐ: ...
\(\Leftrightarrow x^2+\frac{9x^2}{\left(x+3\right)^2}-2x.\frac{3x}{x+3}+\frac{6x^2}{x+3}-40=0\)
\(\Leftrightarrow\left(x-\frac{3x}{x+3}\right)^2+\frac{6x^2}{x+3}-40=0\)
\(\Leftrightarrow\left(\frac{x^2}{x+3}\right)^2+\frac{6x^2}{x+3}-40=0\)
Đặt \(\frac{x^2}{x+3}=t\)
\(\Rightarrow t^2+6t-40=0\Rightarrow\left[{}\begin{matrix}t=4\\t=-10\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\frac{x^2}{x+3}=4\\\frac{x^2}{x+3}=-10\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-4x-12=0\\x^2+10x+30=0\end{matrix}\right.\)
