BÀi 3:
1.a)Thay \(m=\dfrac{1}{2}\) vào pt ta được: \(x^2+x-2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
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b) Để pt có hai nghiệm <=> \(\Delta\ge0\) <=>\(4m^2+8m+4\ge0\) \(\Leftrightarrow4\left(m+1\right)^2\ge0\) (lđ với mọi m)
Theo viet có: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m-1\right)\\x_1x_2=-4m\end{matrix}\right.\)
\(x_1^2+x_2^2-x_1-x_2=6\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2-\left(x_1+x_2\right)=6\)
\(\Leftrightarrow4\left(m-1\right)^2-2.-4m-2\left(m-1\right)=6\)
\(\Leftrightarrow4m^2-2m=0\)
\(\Leftrightarrow\left[{}\begin{matrix}m=0\\m=\dfrac{1}{2}\end{matrix}\right.\)
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