Question

cho phương trình \(x^2-\left(2m-3\right)x+m^2-3m\)=0

Tìm m để pt có 2 nghiệm x1,x2 dể 1<x1<x2

Answer 1

phương trình có 2 nghiệm 1<x1<x2 <=>

\(\left\{{}\begin{matrix}\Delta>0\\a\cdot f\left(1\right)>0\\\dfrac{S}{2}>1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\Delta=9>0\\1-\left(2m-3\right)+m^2-3m>0\\2m-3>1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m^2-5m+4>0\\m>2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}m\in\left(-\infty;1\right)\cup\left(4;+\infty\right)\\m>2\end{matrix}\right.\)

\(\Leftrightarrow m\in\left(4;+\infty\right)\)