a: Thay k=-3 vào pt, ta được:

\(x^2-2\cdot\left(-3+2\right)x+\left(-3\right)^2+2\cdot\left(-3\right)-7=0\)

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

\(\Leftrightarrow\left(x+1\right)^2=5\)

hay \(x\in\left\{\sqrt{5}-1;-\sqrt{5}-1\right\}\)

b: \(\text{Δ}=\left(2k+4\right)^2-4\left(k^2+2k-7\right)\)

\(=4k^2+16k+16-4k^2-8k+28\)

=8k+44

Để phương trình có hai nghiệm thì 8k+44>=0

=>8k>=-44

hay k>=-11/2

Theo đề, ta có: \(\left(x_1+x_2\right)^2-3x_1x_2=28\)

\(\Leftrightarrow\left(2k+4\right)^2-3\cdot\left(k^2+2k-7\right)=28\)

\(\Leftrightarrow4k^2+16k+16-3k^2-6k+21=28\)

\(\Leftrightarrow k^2+10k+37-28=0\)

\(\Leftrightarrow\left(k+1\right)\left(k+9\right)=0\)

=>k=-1

△'=(-2)²-1(m-1)

   =4-m+1

   =5-m

Để PT có 2 no pb thì △'>0

⇒5-m>0

⇒m<5

theo vi-ét ta có

\(\left\{{}\begin{matrix}x_1+x_2=4\\x_1x_2=m-1\end{matrix}\right.\)

mà: \(x^2_1x_2+x_1x_2^2-2\left(x_1+x_2\right)=0\)

⇔\(\left(x_1x_2\right)\left(x_1+x_2\right)-2\left(x_1+x_2\right)=0\)

⇔\(\left(m-1\right)4-2\cdot4=0\)

⇔\(4m-4-8=0\)

⇔4m-12=0

⇔4m=12

⇔m=3

Vậy ...

Lời giải:

Để pt có 2 nghiệm $x_1,x_2$ thì:

$\Delta'=1-(m+2)\geq 0\Leftrightarrow m\leq -1$

Áp dụng định lý Viet:

$x_1+x_2=2$

$x_1x_2=m+2$
Khi đó:
\(\text{VT}=\sqrt{[(x_1-2)^2+mx_2][(x_2-2)^2+mx_1]}=\sqrt{[(x_1-x_1-x_2)^2+mx_2][(x_2-x_1-x_2)^2+mx_1]}\)

\(=\sqrt{(x_2^2+mx_2)(x_1^2+mx_1)}=\sqrt{x_1x_2(x_2+m)(x_1+m)}\)

\(=\sqrt{x_1x_2[x_1x_2+m(x_1+x_2)+m^2]}\)

\(=\sqrt{(m+2)[m+2+2m+m^2]}=\sqrt{(m+2)(m^2+3m+2)}\)

\(=\sqrt{(m+2)^2(m+1)}\)

Lại có:

\(\text{VP}=|x_1-x_2|\sqrt{x_1x_2}=\sqrt{(x_1-x_2)^2x_1x_2}=\sqrt{[(x_1+x_2)^2-4x_1x_2]x_1x_2}\)

\(=\sqrt{-4(m+1)(m+2)}\)

YCĐB thỏa mãn khi:

$\sqrt{(m+1)(m+2)^2}=\sqrt{-4(m+1)(m+2)}$

$\Leftrightarrow (m+1)(m+2)^2=-4(m+1)(m+2)$

$\Leftrightarrow m=-1; m=-2$ hoặc $m=-6$ (đều tm)

Δ=(m+2)^2-4*2m

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

=(m-2)^2>=0

Để PT luôn có hai nghiệm phân biệt thì m-2<>0

=>m<>2

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

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

=>m^2+4m+4-2m-3<=0

=>m^2+2m+1<=0

=>(m+1)^2<=0

=>m=-1

\(x^2-\left(m+3\right)x-m+5=0\)

Theo Vi-ét, ta có :

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=m+3\\x_1x_2=\dfrac{c}{a}=-m+5\end{matrix}\right.\)

Ta có :

\(x_1^2x_2+x_1x_2^2=7\)

\(\Leftrightarrow x_1x_2\left(x_1+x_2\right)-7=0\)

\(\Leftrightarrow\left(-m+5\right)\left(m+3\right)-7=0\)

\(\Leftrightarrow-m^2-3m+5m+15-7=0\)

\(\Leftrightarrow-m^2+2m+8=0\)

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

Ptr có: `a+b+c=1-2m+2+2m-3=0`

   `=>[(x=1),(x=c/a=2m-3):}`

`@TH1: x_1=1;x_2=2m-3`

  `=>\sqrt{1}=2\sqrt{2m-3}`

`<=>\sqrt{2m-3}=1/2`

`<=>2m-3=1/4`

`<=>m=13/8`

`@TH2:x_1=2m-3;x_2=1`

  `=>\sqrt{2m-3}=2\sqrt{1}`

`<=>2m-3=4`

`<=>m=7/2`