\(A=\frac{\sqrt{10}+2\sqrt{6}+\sqrt{10}.\sqrt{4+\sqrt{15}}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}\)

\(A=\frac{\sqrt{10}+2\sqrt{6}+\sqrt{40+10\sqrt{15}}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}\)

\(A=\frac{\sqrt{10}+2\sqrt{6}+\sqrt{\left(5+\sqrt{15}\right)^2}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}\)

\(A=\frac{\sqrt{4}+\sqrt{6}+\sqrt{10}+\sqrt{6}+\sqrt{9}+\sqrt{15}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}\)

\(A=\frac{\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{5}\right)+\sqrt{3}\left(\sqrt{2}+\sqrt{3}+\sqrt{5}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{5}}\)

\(A=\frac{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}+\sqrt{5}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{5}}\)

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

A = \(\frac{\sqrt{10}+2\sqrt{6}+5+\sqrt{15}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}\)

A= \(\frac{\left(\sqrt{2}^2+2\sqrt{2}\sqrt{3}+\sqrt{3}^2\right)+\sqrt{10}+\sqrt{15}}{MC}\)

A= \(\frac{\left(\sqrt{2}+\sqrt{3}\right)^2+\sqrt{5}\left(\sqrt{2}+\sqrt{3}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{5}}\)

A= \(\frac{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}+\sqrt{5}\right)}{\left(\sqrt{2}+\sqrt{3}+\sqrt{5}\right)}\)

A= \(\sqrt{2}+\sqrt{3}\)

cách nào ngắn bạn làm nhé:)) ( cười khinh thk ah t ) 

câu trả lời của t đâu mất rồi. 

Ta có: \(C=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

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

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

Ta có: \(B=\dfrac{\sqrt{2-\sqrt{3}}+\sqrt{4-\sqrt{15}}+\sqrt{10}}{\sqrt{23-3\sqrt{5}}}\)

\(=\dfrac{\sqrt{4-2\sqrt{3}}+\sqrt{8-2\sqrt{15}}+2\sqrt{5}}{3\sqrt{5}-1}\)

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

=1

a) \(\left(\sqrt{14}+\sqrt{6}\right)\sqrt{5-\sqrt{21}}\)

\(=\sqrt{14}\cdot\sqrt{5-\sqrt{21}}+\sqrt{6}\cdot\sqrt{5-\sqrt{21}}\)

\(=\sqrt{14\cdot\left(5-\sqrt{21}\right)}+\sqrt{6\cdot\left(5-\sqrt{21}\right)}\)

\(=\sqrt{70-14\sqrt{21}}+\sqrt{30-6\sqrt{21}}\)

\(=\sqrt{7^2-2\cdot7\cdot\sqrt{21}+\left(\sqrt{21}\right)^2}+\sqrt{\left(\sqrt{21}\right)^2-2\cdot3\cdot\sqrt{21}+3^2}\)

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

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

\(=7-\sqrt{21}+\sqrt{21}-3\)

\(=4\)

b) \(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)

\(=\left[4\cdot\left(\sqrt{10}-\sqrt{6}\right)+\sqrt{15}\cdot\left(\sqrt{10}-\sqrt{6}\right)\right]\cdot\sqrt{4-\sqrt{15}}\)

\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{150}-\sqrt{90}\right)\sqrt{4-\sqrt{15}}\)

\(=\left(4\sqrt{10}-4\sqrt{6}+5\sqrt{6}-3\sqrt{10}\right)\sqrt{4-\sqrt{15}}\)

\(=\left(\sqrt{10}+\sqrt{6}\right)\left(\sqrt{4-\sqrt{15}}\right)\)

\(=\sqrt{10\cdot\left(4-\sqrt{15}\right)}+\sqrt{6\cdot\left(4-\sqrt{15}\right)}\)

\(=\sqrt{40-10\sqrt{15}}+\sqrt{24-6\sqrt{15}}\)

\(=\sqrt{5^2-2\cdot5\cdot\sqrt{15}+\left(\sqrt{15}\right)^2}+\sqrt{\left(\sqrt{15}\right)^2-2\cdot3\cdot\sqrt{15}+3^2}\)

\(=\sqrt{\left(5-\sqrt{15}\right)^2}+\sqrt{\left(\sqrt{15}-3\right)^2}\)

\(=\left|5-\sqrt{15}\right|+\left|\sqrt{15}-3\right|\)

\(=5-\sqrt{15}+\sqrt{15}-3\)

\(=2\)