a) \(\sqrt{15+2\sqrt{5}-\sqrt{21-4\sqrt{5}}}\)

\(=\sqrt{15+2\sqrt{5}-\sqrt{\left(1-2\sqrt{5}\right)^2}}\)

\(=\sqrt{15+2\sqrt{5}-\left(2\sqrt{5}-1\right)}\)

\(=\sqrt{15+2\sqrt{5}-\left(2\sqrt{5}-1\right)}\)

\(=\sqrt{15+2\sqrt{5}-2\sqrt{5}+1}\)

\(=\sqrt{16}\)

\(=4\)

b) \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)

\(=\sqrt[4]{5-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)

\(=\sqrt[4]{5-\sqrt{3-\sqrt{\left(3-2\sqrt{5}\right)^2}}}\)

\(=\sqrt[4]{5-\sqrt{3-\left(2\sqrt{5}-3\right)}}\)

\(=\sqrt[4]{5-\sqrt{3-2\sqrt{5}+3}}\)

\(=\sqrt[4]{5-\sqrt{6-2\sqrt{5}}}\)

\(=\sqrt[4]{5-\sqrt{\left(1-\sqrt{5}\right)^2}}\)

\(=\sqrt[4]{5-\left(\sqrt{5}-1\right)}\)

\(=\sqrt[4]{5-\sqrt{5}+1}\)

\(=\sqrt[4]{6-\sqrt{5}}\)

\(\frac{\sqrt{\sqrt{4+\sqrt{15}}+\sqrt{5-\sqrt{21}}}}{\sqrt{6+\sqrt{35}}}\)+\(\sqrt{\frac{1}{4-2\sqrt{3}}}\)-\(\sqrt{\frac{1}{4+2\sqrt{3}}}\)

=\(\frac{\sqrt{\sqrt{\frac{1}{2}\left(8+2\sqrt{15}\right)}+\sqrt{\frac{1}{2}\left(10-2\sqrt{21}\right)}}}{\sqrt{\frac{1}{2}\left(12+2\sqrt{35}\right)}}\)+\(\sqrt{\frac{1}{3-2\sqrt{3}.1+1}}\)-\(\sqrt{\frac{1}{3+2\sqrt{3}.1+1}}\)

=\(\frac{\sqrt{\sqrt{\frac{1}{2}\left(5+2\sqrt{5}.\sqrt{3}+3\right)}+\sqrt{\frac{1}{2}\left(7-2\sqrt{7}.\sqrt{3}+3\right)}}}{\sqrt{\frac{1}{2}\left(7+2\sqrt{7}.\sqrt{5}+5\right)}}\)+\(\sqrt{\frac{1}{\left(\sqrt{3}-1\right)^2}}\)-\(\sqrt{\frac{1}{\left(\sqrt{3}+1\right)^2}}\)

=\(\frac{\sqrt{\sqrt{\frac{1}{2}\left(\sqrt{5}+\sqrt{3}\right)^2}+\sqrt{\frac{1}{2}\left(\sqrt{7}-\sqrt{3}\right)^2}}}{\sqrt{\frac{1}{2}\left(\sqrt{7}+\sqrt{5}\right)^2}}\)+\(\frac{1}{\sqrt{3}-1}\)-\(\frac{1}{\sqrt{3}+1}\)

=\(\frac{\sqrt{\sqrt{\frac{1}{2}}.\left(\sqrt{5}+\sqrt{3}\right)+\sqrt{\frac{1}{2}}.\left(\sqrt{7}-\sqrt{3}\right)}}{\sqrt{\frac{1}{2}}.\left(\sqrt{7}+\sqrt{5}\right)}\)+\(\frac{\sqrt{3}+1-\sqrt{3}+1}{3-1}\)

=\(\frac{\sqrt{\sqrt{\frac{1}{2}}.\left(\sqrt{7}+\sqrt{5}\right)}}{\sqrt{\frac{1}{2}}.\left(\sqrt{7}+\sqrt{5}\right)}\)+1

=\(\frac{1}{\sqrt{\sqrt{\frac{1}{2}}.\left(\sqrt{7}+\sqrt{5}\right)}}\)+1

a,

b,

a) Ta có: \(A=\sqrt{8-2\sqrt{15}}\cdot\left(\sqrt{3}+\sqrt{5}\right)-\left(\sqrt{45}-\sqrt{20}\right)\)

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

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

\(=\left|\sqrt{5}-\sqrt{3}\right|\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\)

\(=\left(\sqrt{5}-\sqrt{3}\right)\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\)(Vì \(\sqrt{5}>\sqrt{3}\))

\(=5-3-\sqrt{5}\)

\(=2-\sqrt{5}\)

b) Ta có: \(B=\left(\frac{\sqrt{21}-\sqrt{3}}{\sqrt{7}-1}-\frac{\sqrt{15}-\sqrt{3}}{1-\sqrt{5}}\right)\left(\frac{1}{2}\sqrt{6}-\sqrt{\frac{3}{2}}+3\sqrt{\frac{2}{3}}\right)\)

\(=\left(\frac{\sqrt{3}\left(\sqrt{7}-1\right)}{\sqrt{7}-1}+\frac{\sqrt{3}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right)\left(\sqrt{\frac{3}{2}}-\sqrt{\frac{3}{2}}+\sqrt{6}\right)\)

\(=\sqrt{3}+\sqrt{3}+\sqrt{6}\)

\(=2\sqrt{3}+\sqrt{6}\)

c) Ta có: \(C=2\sqrt{3}+\sqrt{7-4\sqrt{3}}+\left(\sqrt{\frac{1}{3}}-\sqrt{\frac{4}{3}}+\sqrt{3}\right):\sqrt{3}\)

\(=2\sqrt{3}+\sqrt{4-2\cdot2\cdot\sqrt{3}+3}+\sqrt{\frac{1}{3}:3}-\sqrt{\frac{4}{3}:3}+\sqrt{3:3}\)

\(=2\sqrt{3}+\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\frac{1}{9}}-\sqrt{\frac{4}{9}}+\sqrt{1}\)

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

\(=2\sqrt{3}+2-\sqrt{3}+\frac{2}{3}\)(Vì \(2>\sqrt{3}\))

\(=\sqrt{3}+\frac{8}{3}\)

d) Ta có: \(D=\left(\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}\right):\frac{1}{\sqrt{7-4\sqrt{3}}}\)

\(=\left(\frac{\left(5+\sqrt{5}\right)^2+\left(5-\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\right)\cdot\sqrt{4-2\cdot2\cdot\sqrt{3}+3}\)

\(=\frac{25+10\sqrt{5}+5+25-10\sqrt{5}+5}{25-5}\cdot\sqrt{\left(2-\sqrt{3}\right)^2}\)

\(=\frac{60}{20}\cdot\left|2-\sqrt{3}\right|\)

\(=3\cdot\left(2-\sqrt{3}\right)\)(Vì \(2>\sqrt{3}\))

\(=6-3\sqrt{3}\)

\(A=\sqrt{20-2.2\sqrt{5}.1+1}-2\sqrt{5}\)

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

\(A=2\sqrt{5}-1-2\sqrt{5}\)

\(A=-1\)

a,\(\left(5+4\sqrt{2}\right)\left(3+2\sqrt{1+\sqrt{2}}\right)\left(3-2\sqrt{1+\sqrt{2}}\right)\)

=\(\left(5+4\sqrt{2}\right)\left(9-4\left(1+\sqrt{2}\right)\right)\)

=\(\left(5+4\sqrt{2}\right)\left(9-4-4\sqrt{2}\right)\)

=\(\left(5+4\sqrt{2}\right)\left(5-4\sqrt{2}\right)=25-\left(4\sqrt{2}\right)^2\)

=-7

b, \(\sqrt{\frac{9}{4}-\sqrt{2}}=\sqrt{\frac{9-4\sqrt{2}}{4}}=\frac{\sqrt{9-4\sqrt{2}}}{2}=\frac{\sqrt{9-2\sqrt{8}}}{2}=\frac{\sqrt{\left(\sqrt{8}-1\right)^2}}{2}=\frac{\left|\sqrt{8}-1\right|}{2}=\frac{\sqrt{8}-1}{2}\)

So sánh:

1) \(2\sqrt{27}\) và \(\sqrt{147}\)

+ \(2\sqrt{27}\) = \(6\sqrt{3}\)

+ \(\sqrt{147}\) = \(7\sqrt{3}\)

⇒ \(6\sqrt{3}\) < \(7\sqrt{3}\)

Vậy: \(2\sqrt{27}\)< \(\sqrt{147}\)

2) \(2\sqrt{15}\) và \(\sqrt{59}\)

+ \(2\sqrt{15}\) = \(\sqrt{60}\)

⇒ \(\sqrt{60}\) > \(\sqrt{59}\)

Vậy: \(2\sqrt{15}\) > \(\sqrt{59}\)

3) \(2\sqrt{2}-1\) và 2

\(giống\left(-1\right)\left\{{}\begin{matrix}3-1\\2\sqrt{2}-1\end{matrix}\right.\)

So sánh: 3 và \(2\sqrt{2}\)

+ 3 = \(\sqrt{9}\)

+ \(2\sqrt{2}=\sqrt{8}\)

⇒ \(\sqrt{8}\) < \(\sqrt{9}\)

⇒ \(\sqrt{8}\) -1 < \(\sqrt{9}\) -1

⇒ \(2\sqrt{2}\) - 1 < 3 - 1

Vậy: \(2\sqrt{2}-1< 2\)

4) \(\frac{\sqrt{3}}{2}\) và 1

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

⇒ \(\frac{\sqrt{3}}{2}\) < \(\frac{2}{2}\)

Vậy: \(\frac{\sqrt{3}}{2}\) < 1

5) \(\frac{-\sqrt{10}}{2}\) và \(-2\sqrt{5}\)

+ \(-2\sqrt{5}\) = \(\frac{-4\sqrt{5}}{2}\) = \(\frac{-\sqrt{80}}{2}\)

⇒ \(\frac{-\sqrt{10}}{2}\) > \(\frac{-\sqrt{80}}{2}\)

Vậy: \(\frac{-\sqrt{10}}{2}\) > \(-2\sqrt{5}\)

\(B=\left(\dfrac{4}{1-\sqrt{5}}+\dfrac{1}{2+\sqrt{5}}-\dfrac{4}{3-\sqrt{5}}\right)\left(\sqrt{5}-6\right)\)

\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)}+\dfrac{2-\sqrt{5}}{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}-\dfrac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\right]\left(\sqrt{5}-6\right)\)

\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{1-5}+\dfrac{2-\sqrt{5}}{4-5}-\dfrac{4\left(3+\sqrt{5}\right)}{9-5}\right]\left(\sqrt{5}-6\right)\)

\(B=\left[-\dfrac{4\left(1+\sqrt{5}\right)}{4}-\dfrac{2-\sqrt{5}}{1}-\dfrac{4\left(3+\sqrt{5}\right)}{4}\right]\left(\sqrt{5}-6\right)\)

\(B=\left(-1-\sqrt{5}-2+\sqrt{5}-3-\sqrt{5}\right)\left(\sqrt{5}-6\right)\)

\(B=\left(-\sqrt{5}-6\right)\left(\sqrt{5}-6\right)\)

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

\(B=-\left(5-36\right)\)

\(B=-\left(-31\right)\)

\(B=31\)

_____________________________

\(\sqrt{48}-\dfrac{\sqrt{21}-\sqrt{15}}{\sqrt{7}-\sqrt{5}}+\dfrac{2}{\sqrt{3}+1}\)

\(=4\sqrt{3}-\dfrac{\sqrt{3}\left(\sqrt{7}-\sqrt{5}\right)}{\sqrt{7}-\sqrt{5}}+\dfrac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)

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

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

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