Đặt \(a=\sqrt{x^2+2}\) (Điều kiện: \(a\ge\sqrt2\) )

=>\(x^2+2=a^2\)

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

Phương trình sẽ trở thành: \(a^2-2+\left(3-a\right)\cdot x=1+2a\)

=>\(a^2-2+3x-a\cdot x-2a-1=0\)

=>\(a^2-a\left(x+2\right)+3x-3=0\) (1)

\(\Delta=\left(x+2\right)^2-4\cdot1\cdot\left(3x-3\right)\)

\(=x^2+4x+4-12x+12=x^2-8x+16=\left(x-4\right)^2\ge0\forall x\)

Do đó: (1) có hai nghiệm phân biệt là:

\(\left[\begin{array}{l}a=\frac{x+2-\left(x-4\right)}{2\cdot1}=\frac{x+2-x+4}{2}=\frac62=3\\ a=\frac{x+2+\left(x-4\right)}{2}=\frac{2x-2}{2}=x-1\end{array}\right.\)

TH1: a=x-1

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

=>\(x^2+2=x^2-2x+1\) và x>=1

=>2=-2x+1 và x>=1

=>2x=-1 và x>=1

=>x=-1/2 và x>=1

=>x∈∅

TH2: a=3

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

=>\(x^2=7\)

=>\(\left[\begin{array}{l}x=\sqrt7\left(nhận\right)\\ x=-\sqrt7\left(nhận\right)\end{array}\right.\)

a: \(\left(x+1\right)\left(-x^2+3x-2\right)<0\)

=>\(\left(x+1\right)\left(x^2-3x+2\right)>0\)

TH1: x+1>0 và \(x^2-3x+2>0\)

=>x>-1 và (x-2)*(x-1)>0

=>x>-1 và (x>2 hoặc x<1)

=>x>2

TH2: x+1<0 và \(x^2-3x+2<0\)

=>x<-1 và (x-2)(x-1)<0

=>x<-1 và 1<x<2

=>x∈∅

b: ĐKXĐ: x>=4

\(\sqrt{x^2-5x+4}+2\sqrt{x+5}>2\sqrt{x-4}+\sqrt{x^2+4x-5}\)

=>\(\sqrt{\left(x-1\right)\left(x-4\right)}-2\sqrt{x-4}-\sqrt{\left(x+5\right)\left(x-1\right)}+2\sqrt{x+5}\) >0

=>\(\sqrt{x-4}\left(\sqrt{x-1}-2\right)-\sqrt{x+5}\left(\sqrt{x-1}-2\right)>0\)

=>\(\left(\sqrt{x-1}-2\right)\left(\sqrt{x-4}-\sqrt{x+5}\right)>0\)

=>\(\sqrt{x-1}-2<0\)

=>\(\sqrt{x-1}<2\)

=>0<=x-1<4

=>1<=x<5

=>4<=x<5

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

Lời giải:
$\Delta'=(\sqrt{3})^2-(\sqrt{3}-1)(1-\sqrt{3})=7-2\sqrt{3}$

PT có 2 nghiệm:
\(x_1=\frac{-b'+\sqrt{\Delta'}}{a}=\frac{\sqrt{3}+\sqrt{7-2\sqrt{3}}}{1-\sqrt{3}}\)

\(x_2=\frac{-b'-\sqrt{\Delta'}}{a}=\frac{\sqrt{3}-\sqrt{7-2\sqrt{3}}}{1-\sqrt{3}}\)

\(x^2+2\left(2+\sqrt{x-1}\right)=5x\)

\(\Leftrightarrow x^2+4+2\sqrt{x-1}-5x=0\)

\(\Leftrightarrow x^2-5x+2\sqrt{x-1}+4=0\)

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

\(x^2+2\left(2+\sqrt{x-1}\right)=5x\left(1\right)\)

Đk: \(x\ge1\)

\(\left(1\right)\Leftrightarrow x^2-5x+4+2\sqrt{x-1}=0\)

\(\Leftrightarrow\left(x^2-4x+4\right)-\left[\left(x-1\right)-2\sqrt{x-1}+1\right]=0\)

\(\Leftrightarrow\left(x-2\right)^2-\left(\sqrt{x-1}-1\right)^2=0\)

\(\Leftrightarrow\left(x+\sqrt{x-1}-3\right)\left(x-\sqrt{x-1}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\sqrt{x-1}-3=0\left(2\right)\\x-\sqrt{x-1}-1=0\left(3\right)\end{matrix}\right.\)

\(\left(2\right)\Leftrightarrow\left(x-1\right)+\sqrt{x-1}-2=0\)

\(\Leftrightarrow\left(x-1\right)-\sqrt{x-1}+2\sqrt{x-1}-2=0\)

\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x-1}-1\right)+2\left(\sqrt{x-1}-1\right)=0\)

\(\Leftrightarrow\left(\sqrt{x-1}-1\right)\left(\sqrt{x-1}+2\right)=0\)

\(\Leftrightarrow\sqrt{x-1}-1=0\) (vì \(\sqrt{x-1}+2>0\))

\(\Leftrightarrow x=2\left(nhận\right)\)

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

\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x-1}-1\right)=0\)

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

Vậy phương trình (1) có 2 nghiệm là \(x=1\text{v}àx=2\)

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

Đặt \(\sqrt{x^2+1}=t>0\)

\(\Rightarrow t^2-\left(x+3\right)t+3x=0\)

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

\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{x+3+x-3}{2}=x\\t=\dfrac{x+3-x+3}{2}=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+1}=x\left(x\ge0\right)\\\sqrt{x^2+1}=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+1=x^2\left(vô-nghiệm\right)\\x=\pm2\sqrt{2}\end{matrix}\right.\)

ĐK: Với mọi x thuộc R.

Ta có: \(x^2+3x+1=\left(x+3\right)\sqrt{x^2+1}\)

\(\Leftrightarrow\left(x^2+3x+1\right)^2=\left[\left(x+3\right)\sqrt{x^2+1}\right]^2\)

\(\Leftrightarrow x^4+6x^3+11x^2+6x+1=\left(x+3\right)^2\left(x^2+1\right)\)

\(\Leftrightarrow x^4+6x^3+11x^2+6x+1=x^4+6x^3+10x^2+6x+9\)

\(\Leftrightarrow x^2-8=0\)

\(\Leftrightarrow x^2=8\)

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

Vậy....

\(\Leftrightarrow\left(x^2+2\right)\sqrt{x^2+x+1}-2\left(x^2+2\right)+x^3-x^2-5x+6=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2+x-3=0\Rightarrow x=...\\x^2+2=\left(2-x\right)\left(\sqrt{x^2+x+1}+2\right)\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow x^2+2x-2=\left(2-x\right)\sqrt{x^2+x+1}\)

Đặt \(\sqrt{x^2+x+1}=t>0\Rightarrow x^2=t^2-x-1\)

\(\Rightarrow t^2+x-3=\left(2-x\right)t\)

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

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

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

\(\Leftrightarrow t=3-x\)

\(\Leftrightarrow\sqrt{x^2+x+1}=3-x\) (\(x\le3\))

\(\Leftrightarrow x^2+x+1=x^2-6x+9\)

\(\Leftrightarrow x=\dfrac{8}{7}\)