Question

Tính
a)\(\sqrt{4-\sqrt{15}}.\sqrt{4+\sqrt{15}}\)
b)\(\sqrt{7-\sqrt{47}}.\sqrt{14+2\sqrt{47}}\)
c)\(\sqrt{4+\sqrt{10+2\sqrt{5}}}.\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
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Answer 1

\(a.\sqrt{4-\sqrt{15}}.\sqrt{4+\sqrt{15}}\)

\(=\sqrt{\left(4-\sqrt{15}\right)\left(4+\sqrt{15}\right)}=\sqrt{\left(16-15\right)}=\sqrt{1}=1\)

\(b.\sqrt{7-\sqrt{47}}.\sqrt{14+2\sqrt{47}}\)

\(=\sqrt{7-\sqrt{47}}.\sqrt{2\left(7-\sqrt{47}\right)}\)

\(=\sqrt{2\left(7-\sqrt{47}\right)\left(7+\sqrt{47}\right)}=\sqrt{2\left(49-47\right)}=\sqrt{2^2}=\sqrt{4}=2\)

\(c.\sqrt{4+\sqrt{10+2\sqrt{5}}}.\sqrt{4-\sqrt{10+2\sqrt{5}}}\)

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

\(=\sqrt{16-\left(\sqrt{10+2\sqrt{5}}\right)^2}\)

\(=\sqrt{16-10-2\sqrt{5}}=\sqrt{6-2\sqrt{5}}=\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}-1\)