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

\(=\sqrt{\frac{6-2\sqrt{5}}{2}}+\sqrt{\frac{6+2\sqrt{5}}{2}}\)

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

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

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

\(=\frac{\sqrt{5}-1+\sqrt{5}+1}{\sqrt{2}}=\frac{2\sqrt{5}}{\sqrt{2}}=\frac{\sqrt{2}.\sqrt{2}.\sqrt{5}}{\sqrt{2}}=\sqrt{10}\)

a: \(\frac{\sqrt{3-\sqrt5}\cdot\left(3+\sqrt5\right)}{\sqrt{10}+\sqrt2}\)

\(=\frac{\sqrt{6-2\sqrt5}\cdot\left(3+\sqrt5\right)}{\sqrt{20}+\sqrt4}\)

\(=\frac{\sqrt{\left(\sqrt5-1\right)^2}\cdot\left(3+\sqrt5\right)}{2\left(\sqrt5+\sqrt1\right)}\)

\(=\frac{\left(\sqrt5-1\right)^{}\cdot\left(3+\sqrt5\right)}{2\left(\sqrt5+\sqrt1\right)}=\frac{3\sqrt5+5-3-\sqrt5}{2\left(\sqrt5+1\right)}=\frac{2\sqrt5+2}{2\sqrt5+2}=1\)

b: \(\sqrt{8\sqrt3}-\sqrt{25\sqrt{12}}+4\sqrt{\sqrt{192}}\)

\(=2\sqrt{2\sqrt3}-5\sqrt{2\sqrt3}+4\sqrt{\sqrt{64}\cdot\sqrt3}\)

\(=-3\sqrt{2\sqrt3}+4\sqrt{8\sqrt3}\)

\(=-3\sqrt{2\sqrt3}+4\cdot2\sqrt{2\sqrt3}=5\sqrt{2\sqrt3}\)

c: \(\sqrt{2-\sqrt3}\cdot\left(\sqrt5+\sqrt2\right)\)

\(=\frac{\sqrt{4-2\sqrt3}\cdot\left(\sqrt5+\sqrt2\right)}{\sqrt2}=\frac{\left(\sqrt3-1\right)\left(\sqrt5+\sqrt2\right)}{\sqrt2}=\frac{\left(\sqrt6-\sqrt2\right)\left(\sqrt5+\sqrt2\right)}{2}\)

\(=\frac{\sqrt{30}+2\sqrt3-\sqrt{10}-2}{2}\)

d: \(\sqrt{3-\sqrt5}+\sqrt{3+\sqrt5}\)

\(=\frac{\sqrt{6-2\sqrt5}+\sqrt{6+2\sqrt5}}{\sqrt2}=\frac{\sqrt{\left(\sqrt5-1\right)^2}+\sqrt{\left(\sqrt5+1\right)^2}}{\sqrt2}\)

\(=\frac{\sqrt5-1+\sqrt5+1}{\sqrt2}=\frac{2\sqrt5}{\sqrt2}=\sqrt{10}\)

e: Đặt \(A=\sqrt{4+\sqrt{10+2\sqrt5}}+\sqrt{4-\sqrt{10+2\sqrt5}}\)

=>\(A^2=4+\sqrt{10+2\sqrt5}+4-\sqrt{10+2\sqrt5}+2\cdot\sqrt{16-\left(10+2\sqrt5\right)}\)

=>\(A^2=8+2\cdot\sqrt{6-2\sqrt5}\)

=>\(A^2=8+2\cdot\sqrt{\left(\sqrt5-1\right)^2}=8+2\left(\sqrt5-1\right)=6+2\sqrt5=\left(\sqrt5+1\right)^2\)

=>\(A=\sqrt5+1\)

f: \(\left(5+2\sqrt6\right)\left(49-20\sqrt6\right)\cdot\sqrt{5-2\sqrt6}\)

\(=\left(245-100\sqrt6+98\sqrt6-240\right)\cdot\sqrt{\left(\sqrt3-\sqrt2\right)^2}\)

\(=\left(5-2\sqrt6\right)\left(\sqrt3-\sqrt2\right)=\left(\sqrt3-\sqrt2\right)^2\)

g: \(\frac{1}{\sqrt2+\sqrt{2+\sqrt3}}+\frac{1}{\sqrt2-\sqrt{2-\sqrt3}}\)

\(=\frac{\sqrt2}{2+\sqrt{4+2\sqrt3}}+\frac{\sqrt2}{2-\sqrt{4-2\sqrt3}}\)

\(=\frac{\sqrt2}{2+\sqrt{\left(\sqrt3+1\right)^2}}+\frac{\sqrt2}{2-\sqrt{\left(\sqrt3-1\right)^2}}\)

\(=\frac{\sqrt2}{2+\left(\sqrt3+1\right)^{}}+\frac{\sqrt2}{2-\left(\sqrt3-1\right)}\)

\(=\frac{\sqrt2}{2+\sqrt3+1^{}}+\frac{\sqrt2}{2-\sqrt3+1}=\frac{\sqrt2}{3+\sqrt3}+\frac{\sqrt2}{3-\sqrt3}=\frac{\sqrt2\left(3-\sqrt3\right)+\sqrt2\left(3+\sqrt3\right)}{9-3}\)

\(=\frac{3\sqrt2-\sqrt6+3\sqrt2+\sqrt6}{6}=\frac{6\sqrt2}{6}=\sqrt2\)

i: \(\frac{\left(\sqrt5+2\right)^2-8\sqrt5}{2\sqrt5-4}\)

\(=\frac{9+4\sqrt5-8\sqrt5}{2\left(\sqrt5-2\right)}\)

\(=\frac{9-4\sqrt5}{2\left(\sqrt5-2\right)}=\frac{\left(\sqrt5-2\right)^2}{2\left(\sqrt5-2\right)}=\frac{\sqrt5-2}{2}\)

k: \(\sqrt{14-8\sqrt3}-\sqrt{24-12\sqrt3}\)
\(=\sqrt{8-2\cdot2\sqrt2\cdot\sqrt6+6}-\sqrt{6\left(4-2\sqrt3\right)}\)
\(=\sqrt{\left(2\sqrt2-\sqrt6\right)^2}-\sqrt6\left(\sqrt3-1\right)=2\sqrt2-\sqrt6-\sqrt{18}+\sqrt6=2\sqrt2-3\sqrt2=-\sqrt2\)

l: \(\frac{4}{\sqrt3+1}+\frac{1}{\sqrt3-2}+\frac{6}{\sqrt3-3}\)

\(=\frac{4\left(\sqrt3-1\right)}{3-1}-\frac{1\left(2+\sqrt3\right)}{\left(2-\sqrt3\right)\left(2+\sqrt3\right)}-\frac{6\left(3+\sqrt3\right)}{9-3}\)

\(=2\left(\sqrt3-1\right)-\left(2+\sqrt3\right)-\left(3+\sqrt3\right)=2\sqrt3-2-2-\sqrt3-3-\sqrt3\)

=-7

m: \(\left(\sqrt2+1\right)^3-\left(\sqrt2-1\right)^3\)

\(=\left(2\sqrt2+3\cdot2\cdot1+3\cdot\sqrt2\cdot1+1\right)-\left(2\sqrt2-3\cdot2\cdot1+3\cdot\sqrt2\cdot1-1\right)\)

\(=\left(5\sqrt2+7\right)-\left(5\sqrt2-7\right)=14\)

a)\(2\sqrt{5}-\sqrt{125}-\sqrt{80}+\sqrt{605}\)

= \(2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}\)

= \(4\sqrt{5}\)

b) \(\sqrt{15-\sqrt{216}}+\sqrt{33-12\sqrt{6}}\)

= \(\sqrt{3\left(5-2\sqrt{6}\right)}-\sqrt{33-12\sqrt{6}}\)

= \(\sqrt{3\left(5-2\sqrt{6}\right)}-\sqrt{3\left(11-4\sqrt{6}\right)}\)

\(a,2\sqrt{5}-\sqrt{125}-\sqrt{80}+\sqrt{605}\)

\(=2\sqrt{5}-\sqrt{5^2.5}-\sqrt{4^2.5}+\sqrt{11^2.5}\)

\(=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}\)

\(=4\sqrt{5}\)

\(b,\sqrt{15-\sqrt{216}}+\sqrt{33-12\sqrt{6}}\)

\(=\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)

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

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

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

\(=\sqrt{6}\)

√8√3 + 2√25√12 - 4√192
= √2.√4.√3 + 2.5.√4.√3 - 4.√16.√4.√3
= 2√6 + 10.√12 - 16.√12
= 2√6 - 6√12
= 2√6 - 6.√4.√3
= 2√6 - 6.2.√3
= 2√6 - 12√3

\(\sqrt{2}\sqrt{4}\sqrt{3}+2.5\sqrt{4}\sqrt{3}-4\sqrt{16}\sqrt{4}\sqrt{3}=2\sqrt{6}+10\sqrt{12}-16\sqrt{12}=2\sqrt{6}-6\sqrt{12}=2\sqrt{6}-12\sqrt{3}\)
 

a) \(\sqrt{24+8\sqrt{5}}+\sqrt{9-4\sqrt{5}}\)

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

\(=3\sqrt{5}\)

b) \(\sqrt{17-12\sqrt{2}}+\sqrt{9+4\sqrt{2}}\)

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

=2

c) \(\sqrt{6-4\sqrt{2}}+\sqrt{22-12\sqrt{2}}\)

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

\(=2\sqrt{2}\)

a: \(=4\sqrt[3]{2}-9\sqrt[3]{2}++6\sqrt[3]{2}=\sqrt[3]{2}\)

b: \(=6\sqrt[3]{3}-15\sqrt[3]{3}+16\sqrt[3]{3}=7\sqrt[3]{3}\)

c: \(=-7\sqrt[3]{3}+3\sqrt[3]{3}+6\sqrt[3]{3}=2\sqrt[3]{3}\)

d: \(=8\sqrt[3]{5}-10\sqrt[3]{5}+2=-2\sqrt[3]{5}+2\)

1: \(\sqrt{3+\sqrt{5}}\cdot\sqrt{2}=\sqrt{6+2\sqrt{5}}=\sqrt{5}+1\)

3) \(\left(\sqrt{\dfrac{3}{4}}-\sqrt{3}+5\cdot\sqrt{\dfrac{4}{3}}\right)\cdot\sqrt{12}\)

\(=\left(\dfrac{\sqrt{3}}{2}-\dfrac{2\sqrt{3}}{2}+5\cdot\dfrac{2}{\sqrt{3}}\right)\cdot\sqrt{12}\)

\(=\dfrac{17\sqrt{3}}{6}\cdot2\sqrt{3}\)

\(=\dfrac{34\cdot3}{6}=\dfrac{102}{6}=17\)