Δ=(-m)^2-4(2m-4)

=m^2-8m+16=(m-4)^2>=0

=>Phương trình luôn có hai nghiệm

a: x1^2+x2^2=13

=>(x1+x2)^2-2x1x2=13

=>m^2-2(2m-4)-13=0

=>m^2-4m-5=0

=>m=5 hoặc m=-1

b: x1^3+x2^3=9

=>(x1+x2)^3-3*x1x2(x1+x2)=9

=>m^3-3*(2m-4)*m=9

=>m^3-6m^2+12m-9=0

=>m=3

\(x^2-\left(m-1\right)x-2=0\)

a=1; b=-m+1; c=-2

Vì a*c=-2<0

nên phương trình luôn có hai nghiệm phân biệt

Theo Vi-et, ta có:

\(\left\{{}\begin{matrix}x_1+x_2=\dfrac{-b}{a}=\dfrac{-\left[-\left(m-1\right)\right]}{1}=m-1\\x_1\cdot x_2=\dfrac{c}{a}=\dfrac{-2}{1}=-2\end{matrix}\right.\)

\(\left(x_1-x_2\right)^2=\left(x_1+x_2\right)^2-4x_1x_2\)

\(=\left(m-1\right)^2-4\cdot\left(-2\right)=\left(m-1\right)^2+8\)

=>\(x_1-x_2=\pm\sqrt{\left(m-1\right)^2+8}\)

\(\dfrac{x_1}{x_2}=\dfrac{x_2^2-3}{x_1^2-3}\)

=>\(x_1\left(x_1^2-3\right)=x_2\left(x_2^2-3\right)\)

=>\(x_1^3-x_2^3=3x_1-3x_2\)

=>\(\left(x_1-x_2\right)\left(x_1^2+x_2^2+x_1x_2-3\right)=0\)

=>\(\left(x_1-x_2\right)\left[\left(x_1+x_2\right)^2-x_1x_2-3\right]=0\)

=>\(\left[{}\begin{matrix}x_1-x_2=0\\\left(m-1\right)^2-\left(-2\right)-3=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}\sqrt{\left(m-1\right)^2+8}=0\left(vôlý\right)\\\left(m-1\right)^2-1=0\end{matrix}\right.\)

=>\(\left(m-1\right)^2=1\)

=>\(\left[{}\begin{matrix}m-1=1\\m-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=2\\m=0\end{matrix}\right.\)

\(\Delta=1-4m\ge0\Rightarrow m\le\dfrac{1}{4}\)

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=-1\\x_1x_2=m\end{matrix}\right.\)

\(x_1^2+x_2^2=3\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=3\)

\(\Leftrightarrow\left(-1\right)^2-2m=3\)

\(\Leftrightarrow-2m=2\)

\(\Rightarrow m=-1\) (thỏa mãn)

a) Khi m = 0 thì phương trình trở thành:

\(x^2+2\left(0-2\right)x-0^2=0\)

\(\Leftrightarrow x^2+2\cdot-2x-0=0\)

\(\Leftrightarrow x^2-4x=0\)

\(\Leftrightarrow x\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

b) Ta có: 

\(\left|x_1\right|-\left|x_2\right|=6\)

\(\Leftrightarrow x^2_1+x_2^2-2\left|x_1x_2\right|=36\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2-2\left|x_1x_2\right|=36\)

Mà: \(x_1+x_2=-2\left(m-2\right)=4-2m\)

\(x_1x_2=-m^2\)

\(\Leftrightarrow\left(4-2m\right)^2-2\cdot-m^2-2\cdot m^2=36\)

\(\Leftrightarrow16-16m+4m^2+2m^2-2m^2=36\)

\(\Leftrightarrow\left(4-2m\right)^2=6^2\)

\(\Leftrightarrow\left[{}\begin{matrix}4-2m=6\\4-2m=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2m=-2\\2m=10\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}m=-1\\m=5\end{matrix}\right.\)

\(\text{Δ}=\left[-2\left(m-2\right)\right]^2-4\cdot1\cdot\left(3m-3\right)\)

\(=\left(2m-4\right)^2-4\left(3m-3\right)\)

\(=4m^2-16m+16-12m+12\)

\(=4m^2-28m+28\)

Để phương trình có hai nghiệm thì Δ>=0

=>\(4m^2-28m+28>=0\)

\(\Leftrightarrow4m^2-2\cdot2m\cdot7+49-21>=0\)

=>\(\left(2m-7\right)^2>=21\)

=>\(\left[{}\begin{matrix}2m-7>=\sqrt{21}\\2m-7< =-\sqrt{21}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m>=\dfrac{7+\sqrt{21}}{2}\\m< =\dfrac{7-\sqrt{21}}{2}\end{matrix}\right.\)

\(\left|x_1\right|-\left|x_2\right|=6\)

=>\(\left(\left|x_1\right|-\left|x_2\right|\right)^2=36\)

=>\(x_1^2+x_2^2-2\left|x_1x_2\right|=36\)

=>\(\left(x_1+x_2\right)^2-2x_1x_2-2\left|x_1x_2\right|=36\)

=>\(\left(-2m+4\right)^2-2\left(3m-3\right)-2\left|3m-3\right|=36\)

=>\(4m^2-16m+16-6m+6-6\left|m-1\right|=36\)

=>\(4m^2-22m+22-36=6\left|m-1\right|\)

=>\(6\left|m-1\right|=4m^2-22m-14\)(1)

TH1: m>=1

(1) tương đương với \(4m^2-22m-14=6\left(m-1\right)\)

=>\(4m^2-22m-14-6m+6=0\)

=>\(4m^2-28m-8=0\)

=>\(m^2-7m-2=0\)

=>\(\left[{}\begin{matrix}m=\dfrac{7+\sqrt{57}}{2}\left(nhận\right)\\m=\dfrac{7-\sqrt{57}}{2}\left(loại\right)\end{matrix}\right.\)

TH2: m<1

(1) tương đương với: \(4m^2-22m-14=6\left(1-m\right)\)

=>\(4m^2-22m-14=6-6m\)

=>\(4m^2-16m-20=0\)

=>m^2-4m-5=0

=>(m-5)(m+1)=0

=>\(\left[{}\begin{matrix}m-5=0\\m+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=5\left(loại\right)\\m=-1\left(nhận\right)\end{matrix}\right.\)