Question

Cho pt \(x^2+2\left(m-1\right)x+4m-11=0\). Tìm m để pt có 2 nghiệm phân biệt mà \(\left(2\left(x_1-1\right)^2+6-x_2\right)\left(x_1x_2+11\right)=72\)

Answer 1

\(\Delta'=m^2-6m+12>0\Rightarrow\) pt luôn có 2 nghiệm phân biệt

Do \(x_1\) là nghiệm nên \(x_1^2+2\left(m-1\right)x_1+4m-11=0\)

\(\Leftrightarrow\left(x_1-1\right)^2=12-4m-2mx_1\)

Theo Viet ta có: \(\left\{{}\begin{matrix}x_1+x_2=-2\left(m-1\right)\\x_1x_2=4m-11\end{matrix}\right.\)

\(2\left(12-4m-2mx_1\right)+\left(6-x_2\right)\left(4m-11+11\right)=72\)

\(\Leftrightarrow24-8m-4mx_1+24m-4mx_2=72\)

\(\Leftrightarrow16m-4m\left(x_1+x_2\right)=48\)

\(\Leftrightarrow2m+m\left(m-1\right)=6\)

\(\Leftrightarrow m^2+m-6=0\Rightarrow\left[{}\begin{matrix}m=2\\m=-3\end{matrix}\right.\)

Answer 2

Nguyễn Việt Lâm giúp mk nhá