Question

cho pt: \(x^2+2\left(m-2\right)x-m^2=0\) . Tìm m để pt có 2no dương pb x_1< x₂ t/m: \(|x_1|-|x_2|=6\)

Answer 1

\(\Delta=4\left(m-2\right)^2+4m^2\)

\(A=8m^2-16m+16\)

Để pt có 2 ng₀ dương pb: \(\left\{{}\begin{matrix}\Delta>0\\P=m^2>0\\S=2m-4>0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}m\ne0\\m>2\end{matrix}\right.\)\(\Rightarrow m>2\)

\(\left|x_1\right|-\left|x_2\right|=6\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-4x_1x_2=36\)

\(\Leftrightarrow4\left(m-2\right)^2-4m^2=36\)

\(\Leftrightarrow m=\frac{-5}{4}\left(KTM\right)\)

Vậy ko tồn tại m.