Question

Cho phương trình:\(x^2-\left(m-1\right)x-m=0\) .Tìm m để pt có hai nghiệm \(x_1;x_2\) thõa mãn:

\(x_1\left(3-x_2\right)+20\ge3\left(2-x_1\right)\)

Answer 1

\(\Delta=\left(m-1\right)^2+4m=\left(m+1\right)^2\ge0\) \(\forall m\Rightarrow\) pt luôn có 2 nghiệm

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

\(x_1\left(3-x_2\right)+20\ge3\left(2-x_1\right)\Leftrightarrow x_1\left(3-x_2\right)+20-3\left(2-x_1\right)\ge0\)

\(\Leftrightarrow3\left(x_1+x_2\right)-x_1x_2+14\ge0\)

\(\Rightarrow3\left(m-1\right)+m+14\ge0\)

\(\Rightarrow4m+11\ge0\)

\(\Rightarrow m\ge\frac{-11}{4}\)