Question

\(B=2\sqrt{4+\sqrt{6-2\sqrt{5}}}.\left(\sqrt{10}-\sqrt{2}\right)\)

Tính B

Answer 1

B=\(2\sqrt{4+\sqrt{6-2\sqrt{5}}}.\left(\sqrt{10}-\sqrt{2}\right)\)

<=>\(B=2\sqrt{4+\sqrt{\left(\sqrt{5}-1\right)^2}}.\sqrt{2}\left(\sqrt{5}-1\right)\)

<=> \(B=2.\sqrt{2}\sqrt{4+\left|\sqrt{5}-1\right|}.\left(\sqrt{5}-1\right)\)

<=>\(B=2\sqrt{2}.\sqrt{4+\sqrt{5}-1}.\left(\sqrt{5}-1\right)\)

<=>\(B=2\sqrt{6+2\sqrt{5}}.\left(\sqrt{5}-1\right)=2\sqrt{\left(\sqrt{5}+1\right)^2}.\left(\sqrt{5}-1\right)=2\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)\)

<=>\(B=2\left(5-1\right)=8\)

Answer 2

B=8