Bài 2: 

a: \(x^2-4x+3=0\)

=>x=1 hoặc x=3

\(x_1^2+x_2^2=1^2+3^2=10\)

b: \(\dfrac{1}{x_1+2}+\dfrac{1}{x_2+2}=\dfrac{1}{1}+\dfrac{1}{5}=\dfrac{6}{5}\)

c: \(x_1^3+x_2^3=1^3+3^3=28\)

d: \(x_1-x_2=1-3=-2\)

a: ĐKXĐ: \(x^2-6x+6\ge0\)

=>\(x^2-6x+9-3\ge0\)

=>\(\left(x-3\right)^2-3\ge0\)

=>\(\left(x-3\right)^2\ge3\)

=>\(\left[\begin{array}{l}x-3\ge\sqrt3\\ x-3\le-\sqrt3\end{array}\right.\Rightarrow\left[\begin{array}{l}x\ge\sqrt3+3\\ x\le-\sqrt3+3\end{array}\right.\)

Ta có: \(x^2-6x+9=4\sqrt{x^2-6x+6}\)

=>\(x^2-6x+6-4\cdot\sqrt{x^2-6x+6}+3=0\)

=>\(\left(\sqrt{x^2-6x+6}-3\right)\left(\sqrt{x^2-6x+6}-1\right)=0\)

TH1: \(\sqrt{x^2-6x+6}-3=0\)

=>\(\sqrt{x^2-6x+6}=3\)

=>\(x^2-6x+6=9\)

=>\(x^2-6x-3=0\)

=>\(x^2-6x+9-12=0\)

=>\(\left(x-3\right)^2=12\)

=>\(\left[\begin{array}{l}x-3=2\sqrt3\\ x-3=-2\sqrt3\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\sqrt3+3\left(nhận\right)\\ x=3-2\sqrt3\left(nhận\right)\end{array}\right.\)

TH2: \(\sqrt{x^2-6x+6}-1=0\)

=>\(x^2-6x+6=1\)

=>\(x^2-6x+5=0\)

=>(x-1)(x-5)=0

=>\(\left[\begin{array}{l}x=1\left(nhận\right)\\ x=5\left(nhận\right)\end{array}\right.\)

b: ĐKXĐ: x∈R

\(x^2-x+8-4\sqrt{x^2-x+4}=0\)

=>\(x^2-x+4-4\cdot\sqrt{x^2-x+4}+4=0\)

=>\(\left(\sqrt{x^2-x+4}-2\right)^2=0\)

=>\(\sqrt{x^2-x+4}-2=0\)

=>\(\sqrt{x^2-x+4}=2\)

=>\(x^2-x+4=4\)

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

=>x(x-1)=0

=>x=0 hoặc x=1

c: \(x^2+\sqrt{4x^2-12x+44}=3x+4\)

=>\(x^2-3x-4+2\sqrt{x^2-3x+11}=0\)

=>\(x^2-3x+11+2\sqrt{x^2-3x+11}-15=0\)

=>\(\left(\sqrt{x^2-3x+11}+5\right)\left(\sqrt{x^2-3x+11}-3\right)=0\)

=>\(\sqrt{x^2-3x+11}-3=0\)

=>\(\sqrt{x^2-3x+11}=3\)

=>\(x^2-3x+11=9\)

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

=>(x-1)(x-2)=0

=>x=1(nhận) hoặc x=2(nhận)

a: =>4x-3x=1-2

=>x=-1

b: =>3x=12

=>x=4

c: =>2(x^2-6)=x(x+3)

=>2x^2-12-x^2-3x=0

=>x^2-3x-12=0

=>\(x=\dfrac{3\pm\sqrt{57}}{2}\)

a: =>4x-3x=1-2

=>x=-1

b: =>3x=12

=>x=4

c: =>2(x^2-6)=x(x+3)

=>2x^2-12=x^2+3x

=>x^2-3x-12=0

=>\(x=\dfrac{3\pm\sqrt{57}}{2}\)

1) `x^2+4-2(x-1)=(x-2)^2`

`<=>x^2+4-2x+2=x^2-4x+4`

`<=>-2x+2=-4x`

`<=>2x=-2`

`<=>x=-1`

.

2) ĐKXĐ: `x \ne \pm 3`

`(x+3)/(x-3)-(x-1)/(x+3)=(x^2+4x+6)/(x^2-9)`

`<=>(x+3)^2-(x-1)(x-3)=x^2+4x+6`

`<=>x^2+6x+9-x^2+4x-3=x^2+4x+6`

`<=>10x+6=x^2+4x+6`

`<=>x^2-6x=0`

`<=>x(x-6)=0`

`<=>x=0;x=6`

.

3) ĐKXĐ: `x \ne \pm 3`

`(3x-3)/(x^2-9) -1/(x-3 )= (x+1)/(x+3)`

`<=>(3x-3)-(x+3)=(x+1)(x-3)`

`<=> 2x-6=x^2-2x-3`

`<=>x^2-4x+3=0`

`<=>x^2-x-3x+3=0`

`<=>x(x-1)-3(x-1)=0`

`<=>(x-3)(x-1)=0`

`<=> x=3;x=1`

Vậy...

a, Câu này dễ quá bỏ qua nha :)

b, Ta có : \(\Delta^,=b^{,2}-ac=\left(-2\right)^2-\left(m+1\right)=4-m-1=3-m\)

- Để phương trình có 2 nghiệm phân biết thì \(\Delta^,>0\)

=> \(m< 3\)

- Theo vi ét : \(\left\{{}\begin{matrix}x_1+x_2=-\frac{b}{a}=4\\x_1x_2=\frac{c}{a}=m+1\end{matrix}\right.\)

- Để \(x^2_1+x^2_2=3\left(x_1+x_2\right)\)

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

<=> \(4^2-2\left(m+1\right)=3.4=12\)

<=> \(-2\left(m+1\right)=-4\)

<=> \(m+1=2\)

<=> \(m=1\left(TM\right)\)

Vậy ....

Lời giải:
Áp dụng hệ thức Viet:

$x_1+x_2=\frac{-4}{3}; x_1x_2=\frac{1}{3}$

Khi đó:
\(B=\frac{x_1}{x_2-1}+\frac{x_2}{x_1-1}=\frac{x_1(x_1-1)+x_2(x_2-1)}{(x_1-1)(x_2-1)}\)

\(=\frac{x_1^2+x_2^2-(x_1+x_2)}{x_1x_2-(x_1+x_2)+1}=\frac{(x_1+x_2)^2-2x_1x_2-(x_1+x_2)}{x_1x_2-(x_1+x_2)+1}\) 

\(=\frac{(\frac{-4}{3})^2-2.\frac{1}{3}-\frac{-4}{3}}{\frac{1}{3}-\frac{-4}{3}+1}=\frac{11}{12}\)