a)Thay m=-1 vào pt có:
\(x^2+2x-8=0\) \(\Leftrightarrow x^2-2x+4x-8=0\)\(\Leftrightarrow\left(x-2\right)\left(x+4\right)=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)
b) Để pt có hai nghiệm pb\(\Leftrightarrow\Delta>0\Leftrightarrow4m^2-4\left(m-1\right)^3>0\)\(\Leftrightarrow m^2>\left(m-1\right)^3\)
Theo định lí viet có:\(\left\{{}\begin{matrix}x_1+x_2=2m\\x_1x_2=\left(m-1\right)^3\end{matrix}\right.\)
Giả sử \(x_1=x_2^2\) .Ta có hệ: \(\left\{{}\begin{matrix}x_1+x_2=2m\\x_1=x_2^2\\x_1x_2=\left(m-1\right)^3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x_1+x_2=2m\\x_1=x_2^2\\x_2^3=\left(m-1\right)^3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x_1+x_2=2m\\x_1=\left(m-1\right)^2\\x_2=m-1\end{matrix}\right.\)
\(\Rightarrow\left(m-1\right)^2+m-1=2m\)
\(\Leftrightarrow m^2-3m=0\) \(\Leftrightarrow\left[{}\begin{matrix}m=3\\m=0\end{matrix}\right.\)(tmđk)
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