\(M=\frac{1}{4}\sqrt{16\times2}-2\sqrt{25\times2}+\sqrt{\frac{22}{11}}\)

\(M=\frac{1}{4}\times4\times\sqrt{2}-2\times5\times\sqrt{2}+\sqrt{2}\)

\(M=\sqrt{2}-10\sqrt{2}+\sqrt{2}\)

\(M=-8\sqrt{2}\)

\(M=\frac{1}{4}\sqrt{16\times2}-2\sqrt{25\times2}+\sqrt{\frac{22}{11}}\)

\(M=\frac{1}{4}\times4\times\sqrt{2}-2\times5\times\sqrt{2}+\sqrt{2}\)

\(M=\sqrt{2}-10\sqrt{2}+\sqrt{2}\)

\(M=-8\sqrt{2}\)

a. \(\sqrt{50}-\sqrt{3}.\sqrt{6}+\frac{\sqrt{22}}{\sqrt{11}}=5\sqrt{2}-3\sqrt{2}+\sqrt{2}=3\sqrt{2}\)

b. \(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{\sqrt{2}+1}-\sqrt{7+4\sqrt{3}}=\frac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\frac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-\sqrt{\left(2+\sqrt{3}\right)^2}\)

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

* \(\frac{1}{\sqrt{8}+\sqrt{7}}+\sqrt{175}-\frac{6\sqrt{2}-4}{3-\sqrt{2}}\)\(=\frac{\left(\sqrt{8}-\sqrt{7}\right)}{\left(\sqrt{8}+\sqrt{7}\right)\left(\sqrt{8}-\sqrt{7}\right)}+\sqrt{25.7}-\frac{2\sqrt{2}\left(3-\sqrt{2}\right)}{3-\sqrt{2}}\)

\(=\sqrt{8}-\sqrt{7}+5\sqrt{7}-2\sqrt{2}=2\sqrt{2}+4\sqrt{7}-2\sqrt{2}=4\sqrt{7}\)

** \(\frac{\sqrt{6-\sqrt{11}}}{\sqrt{22}-\sqrt{2}}+\frac{6}{\sqrt{2}}-\frac{3}{\sqrt{2}+1}\)\(=\frac{\sqrt{2}\sqrt{6-\sqrt{11}}}{\sqrt{2}\left(\sqrt{22}-\sqrt{2}\right)}+\frac{6\sqrt{2}}{2}-\frac{3\left(\sqrt{2}-1\right)}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}\)

\(=\frac{\sqrt{12-2\sqrt{11}}}{2\sqrt{11}-2}+3\sqrt{2}-\frac{3\sqrt{2}-3}{1}\)\(=\frac{\sqrt{\left(\sqrt{11}\right)^2-2.\sqrt{11}+1^2}}{2\left(\sqrt{11}-1\right)}+3\sqrt{2}-3\sqrt{2}+3\)

\(=\frac{\sqrt{11}-1}{2\left(\sqrt{11}-1\right)}+3=\frac{1}{2}+3=\frac{7}{2}\).

Cảm ơn bạn :)

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

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

\(=\frac{-5+12\sqrt{2}}{2}\)

để hình như bạn ghi nhầm rồi

\(A=4\sqrt{32}+2\sqrt{50}-8\sqrt{2}-2\sqrt{98}\)

\(=4\sqrt{16.2}+2\sqrt{25.2}-8\sqrt{2}-2\sqrt{49.2}\)

\(=16\sqrt{2}+10\sqrt{2}-8\sqrt{2}-14\sqrt{2}=4\sqrt{2}\)

\(B=\frac{1}{\sqrt{6}+\sqrt{10}}-\frac{1}{\sqrt{6}-\sqrt{10}}\)

\(=\frac{\sqrt{10}-\sqrt{6}}{\left(\sqrt{6}+\sqrt{10}\right)\left(\sqrt{10}-\sqrt{6}\right)}+\frac{\sqrt{6}+\sqrt{10}}{\left(\sqrt{10}-\sqrt{6}\right)\left(\sqrt{6}+\sqrt{10}\right)}\)

\(=\frac{\sqrt{10}-\sqrt{6}}{4}+\frac{\sqrt{10}+\sqrt{6}}{4}\)

\(=\frac{2\sqrt{10}}{4}=\frac{\sqrt{10}}{2}=\sqrt{2,5}\)

A=\(4\sqrt{2}\)

B=\(\frac{\sqrt{10}}{2}\)

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

= \(8\sqrt{2}+2\sqrt{6}\)

b) = \(2\sqrt{3}-4\sqrt{2}-5\sqrt{3}-\sqrt{2}\)

= \(-3\sqrt{3}-5\sqrt{2}\)

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

=\(\frac{2\sqrt{2}+2-2-\sqrt{2}}{2^2-\sqrt{2^2}}\)

=\(\frac{\sqrt{2}}{4-2}\) = \(\frac{\sqrt{2}}{2}\)

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

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

e) = \(8\sqrt{6}+3\sqrt{6}-6\sqrt{6}=5\sqrt{6}\)

f) = \(4\sqrt{3}+9\sqrt{3}-4\sqrt{3}=9\sqrt{3}\)

g) = \(10+5\sqrt{10}-5\sqrt{10}=10\)

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

= \(\frac{9+3\sqrt{3}+3\sqrt{3}+3}{3^2-\sqrt{3^2}}+\frac{9-3\sqrt{3}-3\sqrt{3}+3}{3^2-\sqrt{3^2}}\)

= \(\frac{12+6\sqrt{3}}{9-3}+\frac{12-6\sqrt{3}}{9-3}\)

= \(\frac{12+6\sqrt{3}+12-6\sqrt{3}}{6}\)

= \(\frac{24}{6}=4\)

k) = \(\left(2\sqrt{7}-2\sqrt{3}+\sqrt{7}\right).\sqrt{7}+2\sqrt{21}\)

= \(\left(3\sqrt{7}-2\sqrt{3}\right).\sqrt{7}+2\sqrt{21}\)

= \(21-2\sqrt{21}+2\sqrt{21}=21\)

l) = \(\frac{\left(2\sqrt{3}-\sqrt{6}\right)\left(\sqrt{8}+2\right)}{\left(\sqrt{8}-2\right)\left(\sqrt{8}+2\right)}\)

= \(\frac{4\sqrt{6}+4\sqrt{3}-4\sqrt{3}-2\sqrt{6}}{\sqrt{8^2}-2^2}\)

= \(\frac{2\sqrt{6}}{8-4}=\frac{2\sqrt{6}}{4}=\frac{\sqrt{6}}{2}\)