Bài 2: 

\(x=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)

Ta có: \(P=x^2-2x+2020\)

\(=4+2\sqrt{3}-2\left(\sqrt{3}-1\right)+2020\)

\(=4+2\sqrt{3}-2\sqrt{3}+2+2020\)

=2026

Bài 1: 

\(A=-\dfrac{3}{4}\cdot\sqrt{9-4\sqrt{5}}\cdot\sqrt{\left(-8\right)^2\cdot\left(2+\sqrt{5}\right)^2}\)

\(=\dfrac{-3}{4}\cdot8\cdot\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)\)

=-6

\(a,=\dfrac{\left(\sqrt{5}-2\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)-\left(2\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}\\ =\dfrac{11-3\sqrt{15}-13-3\sqrt{15}}{2}=\dfrac{-2-6\sqrt{15}}{2}=-1-3\sqrt{15}\)

\(b,=x\sqrt{2\left(x+1\right)}+\sqrt{\dfrac{2\left(x+1\right)^2}{x+1}}-\sqrt{\dfrac{16\left(x+1\right)}{2}}\\ =x\sqrt{2\left(x+1\right)}+\sqrt{2\left(x+1\right)}-2\sqrt{2\left(x+1\right)}\\ =\sqrt{2\left(x+1\right)}\left(x+1-2\right)=\left(x-1\right)\sqrt{2\left(x+1\right)}\)

a.\(=\dfrac{\left(\sqrt{5}-2\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}-\dfrac{\left(2\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}\)

\(=\dfrac{5-\sqrt{15}-2\sqrt{15}+6}{5-3}-\dfrac{10+2\sqrt{15}+\sqrt{15}+3}{5-3}\)

=\(\dfrac{11-3\sqrt{15}-13-3\sqrt{15}}{2}=\dfrac{-2-6\sqrt{15}}{2}\)

=\(-1-3\sqrt{15}\)

b.=\(x\sqrt{2\left(x+1\right)}+\left(x+1\right)\sqrt{\dfrac{2\left(x+1\right)}{\left(x+1\right)^2}}-4\sqrt{\dfrac{2\left(x+1\right)}{2^2}}\)

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

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

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

1)\(=\sqrt{\left(\sqrt{5}-2\right)^2}+\sqrt{26^2}=\sqrt{5}-2+26=24-\sqrt{5}\)

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

3) \(=\dfrac{\sqrt{\left(x-1\right)^2}}{x-1}=\dfrac{\left|x-1\right|}{x-1}\)\(=\left[{}\begin{matrix}1\left(x>1\right)\\-1\left(x< 1\right)\end{matrix}\right.\)

4) \(=\sqrt{\left(\sqrt{\dfrac{7}{2}}+\sqrt{\dfrac{1}{2}}\right)^2}-\sqrt{\left(\sqrt{\dfrac{7}{2}}-\sqrt{\dfrac{1}{2}}\right)^2}=\sqrt{\dfrac{7}{2}}+\sqrt{\dfrac{1}{2}}-\sqrt{\dfrac{7}{2}}+\sqrt{\dfrac{1}{2}}=2\sqrt{\dfrac{1}{2}}=\sqrt{2}\)

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

3. \(\dfrac{\sqrt{x^2-2x+1}}{x-1}=\dfrac{\sqrt{x^2-2.x.1+1^2}}{x-1}=\dfrac{\sqrt{\left(x-1\right)^2}}{x-1}=\dfrac{|x-1|}{x-1}=\left[{}\begin{matrix}x-1>0\left(x>1\right)\\x-1< 0\left(x< 1\right)\end{matrix}\right.=\left[{}\begin{matrix}=1\\=\dfrac{x+1}{x-1}\end{matrix}\right.\)

giúp mk với mk cần gấp

a: \(=6+2\sqrt{11}-4+\sqrt{11}=2+3\sqrt{11}\)

b: \(=\dfrac{3x+9\sqrt{x}-2x+4\sqrt{x}}{\left(\sqrt{x}+3\right)\left(x-2\sqrt{x}\right)}\cdot\dfrac{\left(\sqrt{x}+3\right)^2}{\sqrt{x}+13}=\dfrac{\sqrt{x}+3}{x-2\sqrt{x}}\)