Question

Tìm m để phương trình x²-2019x+m-1=0 có nghiệm x₁, x₂ thỏa mãn x₁²+2018x₂²=2019x₁x₂

Answer 1

\(\Delta=2019^2-4m+4\)

\(x_1^2-x_1x_2-2018x_1x_2+2018x_2^2=0\)

\(\Leftrightarrow x_1\left(x_1-x_2\right)-2018x_2\left(x_1-x_2\right)=0\)

\(\Leftrightarrow\left(x_1-x_2\right)\left(x_1-2018x_2\right)=0\)

TH1: \(x_1=x_2\Rightarrow\Delta=0\Rightarrow2019^2-4m+4=0\Rightarrow m=\frac{2019^2+4}{4}\)

TH2: \(x_1=2018x_2\) kết hợp Viet ta có hệ:

\(\left\{{}\begin{matrix}x_1+x_2=2019\\x_1=2018x_2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x_1=2018\\x_2=1\end{matrix}\right.\)

\(x_1x_2=m-1\Rightarrow m-1=2018\Rightarrow m=2019\)