Lời giải
Với m =1 hệ vô nghiêm
với m khác 1
\(\left(m-1\right)^2\left[\left(m+2\right)^2-4m\right]\ge0\)
\(\Rightarrow\left(m-1\right)^4\ge0\) => đk m khác 1
với m khác 1
ta có
\(\left\{{}\begin{matrix}x_1.x_2=\dfrac{m}{\left(m-1\right)^2}\\x_1+x_2=\dfrac{\left(m+2\right)\left(m-1\right)}{\left(m-1\right)^2}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x_1.x_2=\dfrac{1}{m-1}+\dfrac{1}{\left(m-1\right)^2}\\\left(x_1+x_2\right)-1=\dfrac{3}{\left(m-1\right)}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x_1.x_2=\dfrac{1}{m-1}+\dfrac{1}{\left(m-1\right)^2}\\\dfrac{x_1+x_2-1}{3}=\dfrac{1}{\left(m-1\right)}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x_1.x_2-\left(\dfrac{x_1+x_2-1}{3}\right)=\dfrac{1}{\left(m-1\right)^2}\\\left(\dfrac{x_1+x_2-1}{3}\right)^2=\dfrac{1}{\left(m-1\right)^2}\end{matrix}\right.\)
\(\left(\dfrac{x_1+x_2-1}{3}\right)^2+\left(\dfrac{x_1+x_2-1}{3}\right)-x_1x_2=0\)
