Theo hệ thức Viet: \(\left\{{}\begin{matrix}c+d=-3\\cd=1\end{matrix}\right.\)
Gọi phương trình có 2 nghiệm \(x_1=\frac{2c}{d+1}\) và \(x_2=\frac{2d}{c+1}\)
Ta có: \(\left\{{}\begin{matrix}x_1+x_2=\frac{2c}{d+1}+\frac{2d}{c+1}\\x_1x_2=\frac{4cd}{\left(c+1\right)\left(d+1\right)}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x_1+x_2=\frac{2\left(c^2+d^2\right)+2\left(c+d\right)}{cd+\left(c+d\right)+1}=\frac{2\left(c+d\right)^2-4cd+2\left(c+d\right)}{cd+\left(c+d\right)+1}\\x_1x_2=\frac{4cd}{cd+\left(c+d\right)+1}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x_1+x_2=\frac{2.9-4-6}{1-3+1}=-8\\x_1x_2=\frac{4}{1-3+1}=-4\end{matrix}\right.\)
Theo Viet đảo, \(x_1;x_2\) là nghiệm của: \(x^2+8x-4=0\)
