c) Ta có: \(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(=\dfrac{\left(8+2\sqrt{15}\right)\left(8-2\sqrt{15}\right)}{2}\)
\(=\dfrac{64-60}{2}=2\)
f) Ta có: \(\dfrac{\sqrt{3+\sqrt{5}}}{\sqrt{2}}-\dfrac{\sqrt{5}-1}{2}\)
\(=\dfrac{\sqrt{5}+1-\sqrt{5}+1}{2}\)
=1
a) Ta có: \(\left(7\sqrt{48}+3\sqrt{27}-2\sqrt{12}\right):\sqrt{3}\)
\(=\left(28\sqrt{3}+9\sqrt{3}-4\sqrt{3}\right):\sqrt{3}\)
\(=33\sqrt{3}:\sqrt{3}=33\)
b) Ta có: \(\left(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\right)^2\)
\(=4+\sqrt{7}+4-\sqrt{7}-2\cdot\sqrt{16-7}\)
\(=8-2\cdot3=2\)
