1: \(\text{Δ}=\left(2m-2\right)^2-4\left(m^2-3\right)\)
\(=4m^2-8m+4-4m^2+12=-8m+16\)
Để pt có hai nghiệm phân biệt thì -8m+16>0
hay m<2
2: Theo đề, ta có:
\(\left\{{}\begin{matrix}x_1=3x_2\\x_1+x_2=2m-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_2=\dfrac{2m-2}{4}=\dfrac{m-1}{2}\\x_1=\dfrac{3m-3}{2}\end{matrix}\right.\)
Ta có: \(x_1x_2=m^2-3\)
\(\Leftrightarrow3\left(m-1\right)^2=4\left(m^2-3\right)\)
\(\Leftrightarrow4m^2-12=3m^2-6m+3\)
\(\Leftrightarrow m^2+6m-15=0\)
hay \(m\in\left\{-3+2\sqrt{6};-3-2\sqrt{6}\right\}\)
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