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

=>\(x_1-x_2=\dfrac{1}{2}\left(m+19\right)\)

=>\(\left(x_1-x_2\right)^2=\dfrac{1}{4}\left(m+19\right)^2\)

=>\(\left(x_1+x_2\right)^2-4x_1x_2=\dfrac{1}{4}\left(m+19\right)^2\)

=>\(\left(-2\right)^2-4\left(m-1\right)=\dfrac{1}{4}\left(m+19\right)^2\)

=>\(4-4m+4=\dfrac{1}{4}\left(m+19\right)^2\)

=>\(\left(m+19\right)^2=4\left(-4m+8\right)=-16m+32\)

=>\(m^2+38m+361+16m-32=0\)

=>\(m^2+54m+329=0\)

=>\(\left[{}\begin{matrix}m=-7\left(nhận\right)\\m=-47\left(nhận\right)\end{matrix}\right.\)

a, \(m=-8=>x^2-3x-10=0\)

\(\Delta=\left(-3\right)^2-4\left(-10\right)=49>0\)

=>pt có 2 nghiệm phân biệt \(=>\left[{}\begin{matrix}x1=\dfrac{3+\sqrt{49}}{2}=5\\x2=\dfrac{3-\sqrt{49}}{2}=-2\end{matrix}\right.\)

b, pt(1) \(=>\Delta=\left(-3\right)^2-4\left(m-2\right)=9-4m+8=17-4m\)

pt (1) có 2 nghiệm phân biệt x1,x2 khi \(17-4m>0< =>m< \dfrac{17}{4}\)

theo vi ét \(=>\left\{{}\begin{matrix}x1+x2=3\left(1\right)\\x1x2=m-2\end{matrix}\right.\)

\(x1^3-x2^3+9x1x2=81\)

\(=>\left(x1-x2\right)\left(x1^2+x1x2+x2^2\right)+9\left(m-2\right)=81\)

\(=>x1-x2=\dfrac{81-9\left(m-2\right)}{\left[\left(x1+x2\right)^2-x1x2\right]}\)

\(=>x1-x2=\dfrac{99-9m}{\left[3^2-m+2\right]}=\dfrac{99-9m}{11-m}=9\left(2\right)\)

từ (1)(2)=> hệ pt: \(\left\{{}\begin{matrix}x1+x2=3\\x1-x2=9\end{matrix}\right.=>\left\{{}\begin{matrix}x1=6\\x2=-3\end{matrix}\right.\)

\(=>x1x2=6.\left(-3\right)=m-2=>m=-16\left(tm\right)\)

c: Thay m=-2 vào pt, ta được:

\(x^2-2x+1=0\)

hay x=1

f: Thay x=-3 vào pt, ta được:

\(9-3m+m+3=0\)

=>-2m+12=0

hay m=6

Sửa lại 1 nghiệm thành 2 nghiệm nha

Đề là \(x_1+3x_2=5\) phải không nhỉ?

Vì \(a\cdot c=1\cdot\left(-2\right)=-2< 0\)

nên phương trình luôn có hai nghiệm phân biệt

Theo Vi-et, ta có:

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=m\\x_1x_2=\dfrac{c}{a}=-2\end{matrix}\right.\)

Sửa đề: \(x_1^2\cdot x_2+x_1\cdot x_2^2+7>x_1^2+x_2^2+\left(x_1+x_2\right)^2\)

=>\(x_1x_2\left(x_1+x_2\right)+7>\left(x_1+x_2\right)^2-2x_1x_2+\left(x_1+x_2\right)^2\)

=>\(-2m+7>m^2-2\left(-2\right)+m^2\)

=>\(2m^2+4< -2m+7\)

=>\(2m^2+2m-3< 0\)

=>\(\dfrac{-1-\sqrt{7}}{2}< m< \dfrac{-1+\sqrt{7}}{2}\)