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

a: \(\text{Δ}=\left(2m-2\right)^2-4\left(m-3\right)\)

=4m^2-8m+4-4m+12

=4m^2-12m+16

=4m^2-12m+9+7=(2m-3)^2+7>0

=>Phương trình luôn có nghiệm

b: =>(x1+x2)^2-2x1x2=10

=>(2m-2)^2-2(m-3)=10

=>4m^2-8m+4-2m+6-10=0

=>4m^2-10m=0

=>2m(2m-5)=0

=>m=0 hoặc m=5/2

a: \(x^2-2ax+a^2-a+1=0\)

\(\text{Δ}=\left(-2a\right)^2-4\cdot1\cdot\left(a^2-a+1\right)\)

\(=4a^2-4a^2+4a-4\)

=4a-4

Để phương trình có nghiệm kép thì Δ=0

=>4a-4=0

=>4a=4

=>a=1

Thay a=1 vào phương trình, ta được:

\(x^2-2\cdot1\cdot x+1^2-1+1=0\)

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

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

b: Để phương trình có hai nghiệm thì Δ>=0

=>4a-4>=0

=>4a>=4

=>a>=1

Theo Vi-et, ta có:

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

\(x_1^2+2a\cdot x_2=9\)

=>\(x_1^2+x_2\left(x_1+x_2\right)=9\)

=>\(\left(x_1^2+x_2^2\right)+x_1\cdot x_2=9\)

=>\(\left(x_1+x_2\right)^2-x_1x_2=9\)

=>\(\left(2a\right)^2-\left(a^2-a+1\right)=9\)

=>\(4a^2-a^2+a-1-9=0\)

=>\(3a^2+a-10=0\)

=>\(3a^2+6a-5a-10=0\)

=>(a+2)(3a-5)=0

=>\(\left[{}\begin{matrix}a+2=0\\3a-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=-2\left(lọai\right)\\a=\dfrac{5}{3}\left(nhận\right)\end{matrix}\right.\)

a Khi m=-2 \(\Rightarrow x^2+\left(-2-2\right)x+-2+5=0\Leftrightarrow x^2-4x+3=0\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\) b Theo hệ thức Vi-et có :

\(\Rightarrow\left\{{}\begin{matrix}x_1+x_2=2-m\\x_1x_2=m+5\end{matrix}\right.\)

Mà \(\left(x_1+x_2\right)^2-2x_1x_2=x_1^2+x_2^2=10\Rightarrow\left(2-m\right)^2-2\left(m+5\right)=10\Leftrightarrow m^2-4m+4-2m-10=10\Leftrightarrow m^2-6m-16=0\Leftrightarrow m^2+2m-8m-16=0\Leftrightarrow\left(m+2\right)\left(m-8\right)=0\Leftrightarrow\left[{}\begin{matrix}m=-2\\m=8\end{matrix}\right.\)

a) Thay m=-2 vào phương trình, ta được:

\(x^2-4x+3=0\)

\(\Leftrightarrow x^2-x-3x+3=0\)

\(\Leftrightarrow x\left(x-1\right)-3\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

Vậy: Khi m=-2 thì phương trình có hai nghiệm phân biệt là S={1;3}

a) Phương trình đã cho có \(\Delta'=36-6a+a^2=a^2-6a+9+27=\left(a-3\right)^3+27>0\) nên có 2 nghiệm phân biệt với mọi a

b) Theo hệ thức Vi-et ta có \(x_1+x_2=6\Leftrightarrow x_2=6-x_1\)

Ta có \(x_2=x_1^3-8x_1\Leftrightarrow x_1^3-8x_1=6-x_1\Leftrightarrow x_1^3-7x_1-6=0\)

\(\Leftrightarrow x_1^3-x_1-6x_1-6=0\Leftrightarrow x_1\left(x_1-1\right)\left(x_1+1\right)-6\left(x_1+1\right)=0\)

\(\Leftrightarrow\left(x_1+1\right)\left(x_1^2-x_1-6\right)=0\Leftrightarrow\left(x_1+1\right)\left(x_1^2+2x_1-3x_1-6\right)=0\)

\(\Leftrightarrow\left(x_1+1\right)\left[x_1\left(x_1+2\right)-3\left(x_1+2\right)\right]=0\Leftrightarrow\left(x_1+1\right)\left(x_1+2\right)\left(x_1-3\right)=0\)

\(\Leftrightarrow x_1\in\left\{-1;-2;3\right\}\)

*) \(x_1=-1\Leftrightarrow\left(-1\right)^2-6\left(-1\right)+6a-a^2=0\Leftrightarrow a^2-6a-7=0\Leftrightarrow\orbr{\begin{cases}a=-1\\a=7\end{cases}}\)

*) \(x_1=-2\Leftrightarrow\left(-2\right)^2-6\left(-2\right)+6a-a^2=0\Leftrightarrow a^2-6a-16=0\Leftrightarrow\orbr{\begin{cases}a=-2\\a=8\end{cases}}\)

*) \(x_1=3\Leftrightarrow3^2-6\cdot3+6a-a^2=0\Leftrightarrow a^2-6a+9=0\Leftrightarrow a=3\)

Vậy \(a=\left\{-1;-2;3;7;8\right\}\)

Thay m=-1 vào pt ta được: 

\(x^2+4x-5=0\)\(\Leftrightarrow\left(x-1\right)\left(x+5\right)=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)

Có \(ac=-5< 0\) =>Pt luôn có hai nghiệm pb trái dấu

Theo viet có:\(\left\{{}\begin{matrix}x_1+x_2=2\left(m-1\right)\\2x_1-x_2=11\\x_1x_2=-5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x_1+2x_1-11=2\left(m-1\right)\\x_2=2x_1-11\\x_1x_2=-5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x_1=\dfrac{2m+9}{3}\\x_2=\dfrac{4m-15}{3}\\x_1x_2=-5\end{matrix}\right.\)

\(\Rightarrow\left(\dfrac{2m+9}{3}\right)\left(\dfrac{4m-15}{3}\right)=-5\)\(\Leftrightarrow8m^2+6m-90=0\)

\(\Leftrightarrow\left[{}\begin{matrix}m=3\\m=-\dfrac{15}{4}\end{matrix}\right.\)

Vậy...