\(C=\dfrac{\sqrt{10}-\sqrt{5}+2\sqrt{2}+\sqrt{5}-\sqrt{10}-1}{2\sqrt{2}+2+2\sqrt{2}-1+2\sqrt{2}+2}\)

\(=\dfrac{2\sqrt{2}-1}{6\sqrt{2}+3}=\dfrac{9-4\sqrt{2}}{21}\)

\(B=\dfrac{40}{6+2\sqrt{5}+\sqrt{4\sqrt{5}+4}}\)

\(=\dfrac{40}{\left(\sqrt{5}+1\right)^2+2\sqrt{\sqrt{5}+1}}\)

\(=\dfrac{40}{\sqrt{\sqrt{5}+1}\left(\sqrt{\sqrt{5}+1}+2\right)}\)

\(=\dfrac{40\sqrt{\sqrt{5}-1}}{2\left(\sqrt{\sqrt{5}+1}+2\right)}\)

\(=\dfrac{20\left(\sqrt{\sqrt{5}-1}\right)\left(\sqrt{\sqrt{5}+1}-2\right)}{\sqrt{5}+1-4}\)

\(=\dfrac{20\left(\sqrt{\sqrt{5}-1}\right)\left(\sqrt{\sqrt{5}+1}-2\right)}{-3+\sqrt{5}}\)

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

a: \(=2\cdot\sqrt{\dfrac{18-2\sqrt{77}}{4}}-\sqrt{20+6\sqrt{11}}\)

\(=\sqrt{11}-\sqrt{7}-\sqrt{11}-3=-\sqrt{7}-3\)

b: B=\(=\left(\sqrt{13}-1\right)\cdot\sqrt{\dfrac{7+\sqrt{13}}{18}}+\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)

Đặt \(C=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)

\(\Leftrightarrow C^2=4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\cdot\sqrt{16-10-2\sqrt{5}}\)

\(=8+2\left(\sqrt{5}-1\right)=6+2\sqrt{5}\)

=>\(C=\sqrt{5}+1\)

\(B=\left(\sqrt{13}-1\right)\cdot\sqrt{\dfrac{14+2\sqrt{13}}{36}}+\sqrt{5}+1\)

\(=\dfrac{\left(\sqrt{13}-1\right)\left(\sqrt{13}+1\right)}{6}+\sqrt{5}+1\)

=(13-1)/6+căn5+1

=3+căn5

\(=\dfrac{1}{\sqrt{11-2\sqrt{5}.\sqrt{6}}}-\dfrac{3\left(7+2\sqrt{10}\right)}{\left(7-2\sqrt{10}\right)\left(7+2\sqrt{10}\right)}\\ =\dfrac{1}{\sqrt{\left(\sqrt{5}-\sqrt{6}\right)^2}}-\dfrac{3\left(7+2\sqrt{10}\right)}{49-40}\\ =\dfrac{1}{\left|\sqrt{5}-\sqrt{6}\right|}-\dfrac{7+2\sqrt{10}}{3}\\ =\dfrac{1}{\sqrt{6}-\sqrt{5}}-\dfrac{7+2\sqrt{10}}{3}\\ =\dfrac{\sqrt{6}+\sqrt{5}}{6-5}-\dfrac{7+2\sqrt{10}}{3}\\ =\sqrt{6}+\sqrt{5}+\dfrac{7+2\sqrt{10}}{3}\\ =\dfrac{3\sqrt{6}+3\sqrt{5}+7+2\sqrt{10}}{3}\)

\(=\dfrac{1}{\sqrt{6}-\sqrt{5}}+\dfrac{7+2\sqrt{10}}{3}\)

\(=\sqrt{6}+\sqrt{5}+\dfrac{7}{3}+\dfrac{2}{3}\sqrt{10}\)

1/ Tính: \(A=\dfrac{\sqrt{15-10\sqrt{2}}+\sqrt{13+4\sqrt{10}}-\sqrt{11+2\sqrt{10}}}{2\sqrt{3+2\sqrt{2}}+\sqrt{9-4\sqrt{2}}+\sqrt{12+8\sqrt{2}}}=\dfrac{\sqrt{\left(\sqrt{10}-\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}+\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{10}+1\right)^2}}{2\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(2\sqrt{2}-1\right)^2}+\sqrt{\left(2\sqrt{2}+2\right)^2}}=\dfrac{\sqrt{10}-\sqrt{5}+2\sqrt{2}+\sqrt{5}-\sqrt{10}-1}{2\sqrt{2}+2+2\sqrt{2}-1+2\sqrt{2}+2}=\dfrac{2\sqrt{2}-1}{6\sqrt{2}-3}=\dfrac{2\sqrt{2}-1}{3\left(2\sqrt{2}-1\right)}=\dfrac{1}{3}\)

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

a)

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

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

b)

\(\left(2+\dfrac{11-\sqrt{11}}{1-\sqrt{11}}\right)\left(2+\dfrac{\sqrt{11}+11}{\sqrt{11}+1}\right)\\ =\left(2+\dfrac{\sqrt{11}\left(\sqrt{11}-1\right)}{-\left(\sqrt{11}-1\right)}\right)\left(2+\dfrac{\sqrt{11}\left(1+\sqrt{11}\right)}{\sqrt{11}+1}\right)\\ =\left(2-\sqrt{11}\right)\left(2+\sqrt{11}\right)\\ =4-11\\ =-7\)

a: \(=\left(\dfrac{\sqrt{3}\left(2+\sqrt{3}\right)}{2+\sqrt{3}}-\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\right)\left(\sqrt{3}+\sqrt{2}\right)\)

=(căn 3-căn 2)(căn 3+căn 2)

=3-2=1

b: \(=\left(2-\dfrac{\sqrt{11}\left(\sqrt{11}-1\right)}{\sqrt{11}-1}\right)\left(2+\dfrac{\sqrt{11}\left(\sqrt{11}+1\right)}{\sqrt{11}+1}\right)\)

=(2-căn 11)(2+căn 11)

=4-11

=-7

\(\dfrac{1}{\sqrt{11-2\sqrt{30}}}-\dfrac{3}{\sqrt{7-2\sqrt{10}}}+\dfrac{4}{\sqrt{8+4\sqrt{3}}}\)

\(=\dfrac{1}{\sqrt{11-2.\sqrt{6}.\sqrt{5}}}-\dfrac{3}{\sqrt{7-2.\sqrt{5}.\sqrt{2}}}+\dfrac{4}{\sqrt{2\left(4+2\sqrt{3}\right)}}\)

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

\(=\dfrac{1}{\sqrt{6}+\sqrt{5}}-\dfrac{3}{\sqrt{5}+\sqrt{2}}+\dfrac{2\sqrt{2}}{\sqrt{3}+1}\)

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

Ta có: \(\dfrac{\sqrt{3}+\sqrt{11+6\sqrt{2}}-\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{6+2\sqrt{5}}-\sqrt{7+2\sqrt{10}}}\)

\(=\dfrac{\sqrt{3}+3+\sqrt{2}-\sqrt{3}-\sqrt{2}}{\sqrt{2}+\sqrt{5}+1-\sqrt{5}-\sqrt{2}}\)

=3

a) \(\sqrt{\dfrac{1}{8}}\cdot\sqrt{2}\cdot\sqrt{125}\cdot\sqrt{\dfrac{1}{5}}\) = \(\sqrt{\dfrac{1}{8}\cdot2}.\sqrt{125\cdot\dfrac{1}{5}}=\sqrt{\dfrac{1}{4}}.\sqrt{25}=\dfrac{1}{2}\cdot5=2,5\)

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

c) \(\sqrt{11-6\sqrt{2}}.\sqrt{11+6\sqrt{2}}=\sqrt{\left(11-6\sqrt{2}\right)\left(11+6\sqrt{2}\right)}=\sqrt{121-72}=\sqrt{49}=7\)