\(1,\sqrt{4\left(a-4\right)^2}\left(dkxd:a\ge4\right)\)

\(=\sqrt{4}.\sqrt{\left(a-4\right)^2}\)

\(=\sqrt{2^2}.\left|a-4\right|\)

\(=2\left(a-4\right)\)

\(=2a-8\)

\(2,\sqrt{9\left(b-5\right)^2}\left(dkxd:b< 5\right)\)

\(=\sqrt{9}.\sqrt{\left(b-5\right)^2}\)

\(=\sqrt{3^2}.\left|b-5\right|\)

\(=3\left(-b+5\right)\)

\(=-3b+15\)

Bài 20:

a) \(\sqrt{9-4\sqrt{5}}\cdot\sqrt{9+4\sqrt{5}}=\sqrt{81-80}=1\)

b) \(\left(2\sqrt{2}-6\right)\cdot\sqrt{11+6\sqrt{2}}=2\left(\sqrt{2}-3\right)\left(3+\sqrt{2}\right)\)

\(=2\left(2-9\right)=2\cdot\left(-7\right)=-14\)

c: \(\sqrt{2}\cdot\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)

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

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

=2

d) \(\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{6}-\sqrt{2}\right)\left(2+\sqrt{3}\right)\)

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

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

\(=8+4\sqrt{3}-4\sqrt{3}-6\)

=2

Câu A=4

Cách giải:

\(\left(5\sqrt{3}+2\sqrt{12}-\sqrt{75}\right):\sqrt{3}\)

\(=\left(5\sqrt{3}+2\sqrt{4\cdot3}-\sqrt{25\cdot3}\right)\)\(:\sqrt{3}\)

\(=\left(5\sqrt{3}+4\sqrt{3}-5\sqrt{3}\right)\)\(:\sqrt{3}\)

câu 2

\(...=\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{\left(2+\sqrt{5}\right)^2}=\left|2-\sqrt{5}\right|-\left|2+\sqrt{5}\right|=-4\)

câu 1

\(P=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{1}{\sqrt{x}}\right)\)

\(=\left(\frac{\sqrt{x}\left(3-\sqrt{x}\right)+x+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)

\(=\frac{3\sqrt{x}+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\frac{2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

\(=\frac{3}{\left(3-\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}=\frac{-3\sqrt{x}}{2\sqrt{x}+4}\)

\(P< -1\Leftrightarrow\frac{-3\sqrt{x}}{2\sqrt{x}+4}+1< 0\Leftrightarrow-\sqrt{x}+4< 0\Leftrightarrow\sqrt{x}>4\Leftrightarrow x>16\)

\(C=\sqrt{4-2\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)

\(\Leftrightarrow C=\sqrt{3-2\sqrt{3}+1}-\sqrt{4+4\sqrt{3}+3}\)

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

\(\Leftrightarrow C=\left|\sqrt{3}-1\right|-\left|2+\sqrt{3}\right|\)

\(\Leftrightarrow C=\sqrt{3}-1-2-\sqrt{3}\)

\(\Leftrightarrow C=-3\)

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: \(D=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)

\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)

\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)

b: \(E=\sqrt{6-2\sqrt{5}}\cdot\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)

\(=\left(6-2\sqrt{5}\right)\left(3+\sqrt{5}\right)\)

\(=18+6\sqrt{5}-6\sqrt{5}-10=8\)

\(=2\left|3-\sqrt{2}\right|+\sqrt{18}-5.1=6-2\sqrt{2}+3\sqrt{2}-5\)

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