Question

Tìm các gtri của m để ptr \(x^2-2\left(m+1\right)x+4m-3\)=0 có 2 nghiệm x1,x2 thỏa mãn \(x^2_1x_2+x_{_{ }1}x^2_2\)=4

Answer 1

\(x^2-2\left(m+1\right)x+4m-3=0\)

Theo Vi - ét, ta có :

\(\left\{{}\begin{matrix}x_`+x_2=-\dfrac{b}{a}=2\left(m+1\right)=2m+2\\x_1x_2=\dfrac{c}{a}=4m-3\end{matrix}\right.\)

Ta có :

\(x_1^2x_2+x_1x_2^2=4\)

\(\Leftrightarrow x_1x_2\left(x_1+x_2\right)-4=0\)

\(\Leftrightarrow\left(4m-3\right)\left(2m+2\right)-4=0\)

\(\Leftrightarrow8m^2+8m-6m-6-4=0\)

\(\Leftrightarrow8m^2+2m-10=0\)

\(\Leftrightarrow\left[{}\begin{matrix}m=1\\m=-\dfrac{5}{4}\end{matrix}\right.\)