1: \(\Delta=2^2-4\cdot1\left(m-1\right)\)
\(=4-4m+4=-4m+8\)
Để phương trình có hai nghiệm phân biệt thì \(\Delta>0\)
=>-4m+8>0
=>-4m>-8
=>m<2
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=-2\\x_1\cdot x_2=\dfrac{c}{a}=m-1\end{matrix}\right.\)
\(x_1^3+x_2^3-6x_1x_2=4\left(m-m^2\right)\)
=>\(\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)-6x_1x_2=4\left(m-m^2\right)\)
=>\(\left(-2\right)^3-3\cdot\left(-2\right)\left(m-1\right)-6\left(m-1\right)=4\left(m-m^2\right)\)
=>\(-8+6\left(m-1\right)-6\left(m-1\right)=4\left(m-m^2\right)\)
=>\(4\left(m^2-m\right)=8\)
=>\(m^2-m=2\)
=>\(m^2-m-2=0\)
=>(m-2)(m+1)=0
=>\(\left[{}\begin{matrix}m-2=0\\m+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=2\left(loại\right)\\m=-1\left(nhận\right)\end{matrix}\right.\)
2: \(x_1^2+2x_2+2x_1x_2+20=0\)
=>\(x_1^2-x_2\left(x_1+x_2\right)+2x_1x_2+20=0\)
=>\(x_1^2-x_2^2+x_1x_2+20=0\)
=>\(\left(x_1-x_2\right)\left(x_1+x_2\right)+m-1+20=0\)
=>\(-2\left(x_1-x_2\right)=-m-19\)
=>2(x1-x2)=m+19
=>\(x_1-x_2=\dfrac{1}{2}\left(m+19\right)\)
=>\(\left(x_1-x_2\right)^2=\dfrac{1}{4}\left(m+19\right)^2\)
=>\(\left(x_1+x_2\right)^2-4x_1x_2=\dfrac{1}{4}\left(m+19\right)^2\)
=>\(\left(-2\right)^2-4\left(m-1\right)=\dfrac{1}{4}\left(m+19\right)^2\)
=>\(4-4m+4=\dfrac{1}{4}\left(m+19\right)^2\)
=>\(\left(m+19\right)^2=4\left(-4m+8\right)=-16m+32\)
=>\(m^2+38m+361+16m-32=0\)
=>\(m^2+54m+329=0\)
=>\(\left[{}\begin{matrix}m=-7\left(nhận\right)\\m=-47\left(nhận\right)\end{matrix}\right.\)
