Question

1)\(\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3-\sqrt{5}}=8\)

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

Answer 1

1) \(\left(3+\sqrt{5}\right)\sqrt{2}\left(\sqrt{5}-1\right)\sqrt{3-\sqrt{5}}\\ =\sqrt{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}.\sqrt{6+2\sqrt{5}}.\left(\sqrt{5}-1\right)\\ =\sqrt{9-5}.\sqrt{\left(\sqrt{5}+1\right)^2}.\left(\sqrt{5}-1\right)\\ =2.\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)\\ =2\left(5-1\right)=8=VP\)

=> ĐPCM

2) Bình phương 2 về, đpcm <=> \(2\sqrt{2}-2\sqrt{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}\) = \(2\left(\sqrt{2}-1\right)\)

VT = \(2\sqrt{2}-2\sqrt{2-1}=2\left(\sqrt{2}-1\right)\) = VP=> ĐPCM

Answer 2

1) Ta có: \(\left(3+\sqrt{5}\right)\cdot\left(\sqrt{10}-\sqrt{2}\right)\cdot\sqrt{3-\sqrt{5}}\)

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

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

=8