Ta có \(a+b+c=1-2m+2m-1=0\)
\(\Rightarrow\) Phương trình luôn có 2 nghiệm \(\left\{{}\begin{matrix}x_1=1\\x_2=2m-1\end{matrix}\right.\)
Để pt có 2 nghiệm pb đều dương thì:
\(\Leftrightarrow\left\{{}\begin{matrix}2m-1>0\\2m-1\ne1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}m>\frac{1}{2}\\m\ne1\end{matrix}\right.\)
b/ \(A=2x_1^2+x_2^2-2x_1x_2=\left(x_1-x_2\right)^2+x_1^2\)
- Nếu \(x_1=1\) thì \(A=\left(1-x_2\right)^2+1>1>\frac{1}{2}\) BĐT đúng
- Nếu \(\left\{{}\begin{matrix}x_1=2m-1\\x_2=1\end{matrix}\right.\) thì
\(A=2\left(2m-1\right)^2+1-2\left(2m-1\right)\)
\(=2\left[\left(2m-1\right)^2-\left(2m-1\right)+\frac{1}{4}\right]+\frac{1}{2}\)
\(=2\left(2m-1-\frac{1}{2}\right)^2+\frac{1}{2}=2\left(2m-\frac{3}{2}\right)^2+\frac{1}{2}\ge\frac{1}{2}\)
Dấu "=" xảy ra khi \(2m-\frac{3}{2}=0\Leftrightarrow m=\frac{3}{4}\)
