a) \(^{x^2}\)-2(m+1)x +m-4 = 0 (1)
(a=1; b= -2(m+1); c= m-4)
Để pt (1) có 2 nghiệm pb thì:
\(\left\{{}\begin{matrix}ae0\\\Delta>0\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}1e0\left(đ\right)\\b^2-4ac>0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\Delta=[-2\left(m+1\right)]^2-4.1.\left(m-4\right)\)
\(\Leftrightarrow\Delta=4m^2+8m+4-4m+16\)
\(\Leftrightarrow\Delta=4m^2+4m+20\)
\(\Leftrightarrow\Delta=\left(2m+1\right)^2+19>0,\forall m\)
\(\Rightarrow\)Đpcm
b) Theo đl Vi-ét ta có:
\(\left\{{}\begin{matrix}S=x_1+x_2=\dfrac{-b}{a}=2\left(m+1\right)\\P=x_1.x_2=\dfrac{c}{a}=m-4\end{matrix}\right.\)
Ta có \(M=|x_1-x_2|\)
\(\Rightarrow M^2=\left(x_1-x_2\right)^2\)
\(\Rightarrow M^2=x_1^2-2x_1x_2+x_2^2\)
\(\Rightarrow M^2=\left(x_1+x_2\right)^2-4x_1x_2\)
\(\Rightarrow M^2=[2\left(m+1\right)]^2-4\left(m-4\right)\)
\(\Rightarrow M^2=4m^2+8m+4-4m+16\)
\(\Rightarrow M^2=4m^2+4m+20\)
\(\Rightarrow M^2=\left(2m+1\right)^2+19\)
Ta có \(\left(2m+1\right)^2\ge0\)
\(\Rightarrow\left(2m+1\right)^2+19\ge19\)
\(\Rightarrow M^2\ge19\)
\(\Rightarrow M\ge\sqrt{19}\)(vì \(|x_1-x_2|\ge0\))
Dấu "=" xảy ra: \(2m+1=0\Leftrightarrow m=-\dfrac{1}{2}\)
Vậy \(minM=\sqrt{19}\Leftrightarrow m=-\dfrac{1}{2}\)
