Question

1. Cho phương trình: \(x^2+2\left(m-1\right)x+4m-11\). Tìm m để phương trình có 2 nghiệm \(x_1,x_2\)sao cho

\(2\left(x_1-1\right)^2+\left(6-x_2\right)\left(x_1x_2+11\right)=72\)

Answer 1

\(\Delta'=\left(m-1\right)^2-4m+11=\left(m-3\right)^2+3>0\)

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

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

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

Thay vào:

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

\(\Leftrightarrow24-4m-4mx_1+24m-4mx_2-72=0\)

\(\Leftrightarrow-4m\left(x_1+x_2\right)+20m-48=0\)

\(\Leftrightarrow2m\left(m-1\right)+5m-12=0\)

\(\Leftrightarrow...\)