b) Ta có: \(x+\sqrt{3}=2\Leftrightarrow x-2=-\sqrt{3}\Leftrightarrow\left(x-2\right)^2=3\Leftrightarrow x^2-4x+1=0\)

\(B=x^5-3x^4-3x^3+6x^2-20x+2021\)

\(B=\left(x^5-4x^4+x^3\right)+\left(x^4-4x^3+x^2\right)+\left(5x^2-20x+5\right)+2016\)

\(B=x^3\left(x^2-4x+1\right)+x^2\left(x^2-4x+1\right)+5\left(x^2-4x+1\right)+2016\)

Thế \(x^2-4x+1=0\)\(\Rightarrow B=2016.\)

Bài 1: 

a) \(\dfrac{a+\sqrt{a}}{\sqrt{a}}=\sqrt{a}+1\)

b) \(\dfrac{\sqrt{\left(x-3\right)^2}}{3-x}=\dfrac{\left|x-3\right|}{3-x}=\pm1\)

Bài 2: 

a) \(\dfrac{\sqrt{9x^2-6x+1}}{9x^2-1}=\dfrac{\left|3x-1\right|}{\left(3x-1\right)\left(3x+1\right)}=\pm\dfrac{1}{3x+1}\)

b) \(4-x-\sqrt{x^2-4x+4}=4-x-\left|x-2\right|=\left[{}\begin{matrix}6-2x\left(x\ge2\right)\\2\left(x< 2\right)\end{matrix}\right.\)

a)\(pt\Leftrightarrow\sqrt{x^2-2x+2}+\sqrt{3x^2-6x+4}-2=0\)

\(\Leftrightarrow\sqrt{x^2-2x+2}-1+\sqrt{3x^2-6x+4}-1=0\)

\(\Leftrightarrow\frac{x^2-2x+2-1}{\sqrt{x^2-2x+2}+1}+\frac{3x^2-6x+4-1}{\sqrt{3x^2-6x+4}+1}=0\)

\(\Leftrightarrow\frac{x^2-2x+1}{\sqrt{x^2-2x+2}+1}+\frac{3x^2-6x+3}{\sqrt{3x^2-6x+4}+1}=0\)

\(\Leftrightarrow\frac{\left(x-1\right)^2}{\sqrt{x^2-2x+2}+1}+\frac{3\left(x-1\right)^2}{\sqrt{3x^2-6x+4}+1}=0\)

\(\Leftrightarrow\left(x-1\right)^2\left(\frac{1}{\sqrt{x^2-2x+2}+1}+\frac{3}{\sqrt{3x^2-6x+4}+1}\right)=0\)

Dễ thấy: \(\frac{1}{\sqrt{x^2-2x+2}+1}+\frac{3}{\sqrt{3x^2-6x+4}+1}>0\) (loại)

Nên x-1=0 suy ra x=1

b)\(pt\Leftrightarrow\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+21}+x^2+2x-5=0\)

\(\Leftrightarrow\sqrt{3x^2+6x+7}-2+\sqrt{5x^2+10x+21}-4+x^2+2x+1=0\)

\(\Leftrightarrow\frac{3x^2+6x+7-4}{\sqrt{3x^2+6x+7}+2}+\frac{5x^2+10x+21-16}{\sqrt{5x^2+10x+21}+4}+\left(x+1\right)^2=0\)

\(\Leftrightarrow\frac{3\left(x+1\right)^2}{\sqrt{3x^2+6x+7}+2}+\frac{5\left(x+1\right)^2}{\sqrt{5x^2+10x+21}+4}+\left(x+1\right)^2=0\)

\(\Leftrightarrow\left(x+1\right)^2\left(\frac{3}{\sqrt{3x^2+6x+7}+2}+\frac{5}{\sqrt{5x^2+10x+21}+4}+1\right)=0\)

Dễ thấY: \(\frac{3}{\sqrt{3x^2+6x+7}+2}+\frac{5}{\sqrt{5x^2+10x+21}+4}+1>0\) (loại luôn)

Nên x+1=0 suy ra x=-1