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)

=>\(\left(\dfrac{x^2-8}{2008}-1\right)+\left(\dfrac{x^2-7}{2009}-1\right)=\left(\dfrac{x^2-6}{2010}-1\right)+\left(\dfrac{x^2-5}{2011}-1\right)\)

=>x^2-2016=0

=>x^2=2016

=>\(x=\pm\sqrt{2016}\)

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

Giải:

ĐK: \(x\ge-1\)

PT tương đương với: \(\sqrt{x+1}=1-x^2\)

\(\Leftrightarrow\left\{{}\begin{matrix}1-x^2\ge0\\x+1=1-2x^2+x^4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}-1\le x^2\le1\\x^4-2x^2-x=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}-1\le x\le1\\\left\{{}\begin{matrix}x=0\left(TM\right)\\x^3-2x-1=0\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}-1\le x\le1\\\left[{}\begin{matrix}x=0\\x=-1\\x^2-x-1=0\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=\dfrac{1-\sqrt{5}}{2}\end{matrix}\right.\)

2:

\(A=\dfrac{x_2-1+x_1-1}{x_1x_2-\left(x_1+x_2\right)+1}\)

\(=\dfrac{3-2}{-7-3+1}=\dfrac{1}{-9}=\dfrac{-1}{9}\)

B=(x1+x2)^2-2x1x2

=3^2-2*(-7)

=9+14=23

C=căn (x1+x2)^2-4x1x2

=căn 3^2-4*(-7)=căn 9+28=căn 27

D=(x1^2+x2^2)^2-2(x1x2)^2

=23^2-2*(-7)^2

=23^2-2*49=431

D=9x1x2+3(x1^2+x2^2)+x1x2

=10x1x2+3*23

=69+10*(-7)=-1

\(x^2-2x+1< 9\)

\(\Leftrightarrow\left(x-1\right)^2< 9\)

\(\Leftrightarrow x-1< 3\)

\(\Leftrightarrow x< 4\)

\(\left(x-1\right)\left(4-x^2\right)\ge0\)

\(\Leftrightarrow\left(x-1\right)\left(2-x\right)\left(2+x\right)\ge0\)

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

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

\(\dfrac{x+2}{x-5}< 0\)

\(\Leftrightarrow x+2< 0\)

\(\Leftrightarrow x< -2\)

a)\(x^2-2x+1< 9\)

\(\Leftrightarrow\left(x-1\right)^2< 9\)

\(\Leftrightarrow\left(x-1\right)^2-9< 0\)

\(\Leftrightarrow\left(x-1-3\right)\left(x-1+3\right)< 0\)

\(\Leftrightarrow\left(x-4\right)\left(x+2\right)< 0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4< 0\\x+2>0\end{matrix}\right.hay\left[{}\begin{matrix}x-4>0\\x+2< 0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x< 4\\x>-2\end{matrix}\right.hay\left[{}\begin{matrix}x>4\\x< -2\end{matrix}\right.\)(vô lý)

-Vậy nghiệm của BĐT là \(-2< x< 4\).

b) \(\left(x-1\right)\left(4-x^2\right)\ge0\)

\(\Leftrightarrow\left(x-1\right)\left(2-x\right)\left(x+2\right)\ge0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x+2\right)\le0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1< 0\\x-2>0\\x+2>0\end{matrix}\right.\) hay \(\left[{}\begin{matrix}x-1>0\\x-2< 0\\x+2>0\end{matrix}\right.\) hay \(\left[{}\begin{matrix}x-1>0\\x-2 >0\\x+2< 0\end{matrix}\right.\) hay \(\left[{}\begin{matrix}x-1< 0\\x-2< 0\\x+2< 0\end{matrix}\right.\)

 \(\Leftrightarrow\left[{}\begin{matrix}x< 1\\x>2\\x>-2\end{matrix}\right.\) (vô lí) hay \(\left[{}\begin{matrix}x>1\\x< 2\\x>-2\end{matrix}\right.\) (có thể xảy ra) hay

\(\left[{}\begin{matrix}x>1\\x>2\\x< -2\end{matrix}\right.\) (vô lí) hay \(\left[{}\begin{matrix}x< 1\\x< 2\\x< -2\end{matrix}\right.\) (có thể xảy ra)

-Vậy nghiệm của BĐT là \(x< -2\) hay \(1< x< 2\).

c) ĐKXĐ: \(x\ne5\)

 \(\dfrac{x+2}{x-5}< 0\Leftrightarrow\left[{}\begin{matrix}x+2< 0\\x-5>0\end{matrix}\right.hay\left[{}\begin{matrix}x+2>0\\x-5< 0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< -2\\x>5\end{matrix}\right.\)(vô lí) hay

\(\left[{}\begin{matrix}x>-2\\x< 5\end{matrix}\right.\) (có thể xảy ra)

-Vậy nghiệm của BĐT là \(-2< x< 5\)

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

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

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

=>x=3 hoặc x=-2

     x²-9=(x-3)(2x-5)
(=) (x-3)(x+3)=(x-3)(2x-5)

(=) (x-3)(x+3)-(x-3)(2x-5)=0

(=) (x-3)(x+3-2x+5)=0
(=) (x-3)(8-x)=0

(=)x-3=0 hoặc 8-x=0

(=)x=0 hoặc x=8

Vậy S=\(\left\{0;8\right\}\)

a: =>(5x+3)(x-1)=0

=>x=1 hoặc x=-3/5

b: =>(x-3)(4x-1-5x-2)=0

=>(x-3)(-x-3)=0

=>x=-3 hoặc x=3

c: =>(x+6)(3x-1+x-6)=0

=>(x+6)(4x-7)=0

=>x=7/4 hoặc x=-6

Bài 1: 

a: \(\Leftrightarrow x^2-5x+6< =0\)

=>(x-2)(x-3)<=0

=>2<=x<=3

b: \(\Leftrightarrow\left(x-6\right)^2< =0\)

=>x=6

c: \(\Leftrightarrow x^2-2x+1>=0\)

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

hay \(x\in R\)

tham khảo 

https://hoidapvietjack.com/q/57243/giai-cac-phuong-trinh-sau-a-2x12-2x-12-b-x2-3x-2-5x2-3x60

b) (2x+1)²-2x-1=2

\(< =>4x^2+4x+1-2x-1=2\)

\(< =>4x^2+2x-2=0\)

\(< =>4x^2+4x-2x-2=0\)

\(< =>\left(4x^2+4x\right)-\left(2x+2\right)=0\)

\(< =>4x\left(x+1\right)-2\left(x+1\right)=0\)

\(< =>\left(x+1\right)\left(4x-2\right)=0\)

\(=>\left\{{}\begin{matrix}x+1=0=>x=-1\\4x-2=0=>x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy....