Question

[Jessica Võ]

\(A=\left(5+\frac{7-\sqrt{21}}{\sqrt{7}-\sqrt{3}}\right)\left(\sqrt{16-5\sqrt{7}}\right)\)

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Answer 1

[Lê Gia Bảo]

Ta có: \(A=\left(5+\frac{7-\sqrt{21}}{\sqrt{7}-\sqrt{3}}\right)\sqrt{16-5\sqrt{7}}\)

\(=\left(5+\frac{\sqrt{7}\left(\sqrt{7}-\sqrt{3}\right)}{\sqrt{7}-\sqrt{3}}\right)\sqrt{16-5\sqrt{7}}\)

\(=\left(5+\sqrt{7}\right)\sqrt{16-5\sqrt{7}}\)

\(\Rightarrow\sqrt{2}A=\left(5+\sqrt{7}\right)\sqrt{32-10\sqrt{7}}\)

\(=\left(5+\sqrt{7}\right)\sqrt{\left(5-\sqrt{7}\right)^2}=\left(5+\sqrt{7}\right)\left(5-\sqrt{7}\right)\)

\(=25-7=18\)

Vậy \(A=\frac{18}{\sqrt{2}}=9\sqrt{2}\)