Question

giúp mình với ạ.

Answer 1

a: \(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)^2\)

\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)

\(=32-8\sqrt{15}+8\sqrt{15}-30\)

=2

b: \(=\dfrac{1}{\sqrt{2}}\cdot\left[\left(3-\sqrt{5}\right)\left(\sqrt{5}+1\right)+\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\right]\)

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

\(=\dfrac{1}{\sqrt{2}}\cdot4\sqrt{5}=2\sqrt{10}\)

c: \(=\dfrac{1}{\sqrt{2}}\left(\sqrt{5}+1-\sqrt{5}+1\right)-\sqrt{2}=\sqrt{2}-\sqrt{2}=0\)

d: \(=\dfrac{1}{\sqrt{2}}\left(\sqrt{7}-1-\sqrt{7}-1\right)+\sqrt{7}=-\sqrt{2}+\sqrt{7}\)

e: \(=\dfrac{1}{\sqrt{2}}\left(\sqrt{12}+1+\sqrt{12}-1\right)+2\sqrt{6}=2\sqrt{6}+2\sqrt{6}=4\sqrt{6}\)