Question

Tính \(x=\sqrt{4+\sqrt{8}}.\sqrt{2+\sqrt{2+\sqrt{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}\)

\(y=\dfrac{3\sqrt{8}-2\sqrt{12}+\sqrt{20}}{3\sqrt{18}-2\sqrt{27}+\sqrt{45}}\)

Answer 1

Lời giải:

\(x=\sqrt{4+\sqrt{8}}.\sqrt{(2+\sqrt{2+\sqrt{2}})(2-\sqrt{2+\sqrt{2}})}\)

\(=\sqrt{4+\sqrt{8}}.\sqrt{2^2-(2+\sqrt{2})}=\sqrt{4+\sqrt{8}}.\sqrt{2-\sqrt{2}}\)

\(=\sqrt{2(2+\sqrt{2})}.\sqrt{2-\sqrt{2}}=\sqrt{2}.\sqrt{(2+\sqrt{2})(2-\sqrt{2})}\)

\(=\sqrt{2}.\sqrt{2^2-2}=\sqrt{2}.\sqrt{2}=2\)

\(y=\frac{3.2\sqrt{2}-2.2\sqrt{3}+2\sqrt{5}}{3.3\sqrt{2}-2.3\sqrt{3}+3\sqrt{5}}=\frac{6\sqrt{2}-4\sqrt{3}+2\sqrt{5}}{9\sqrt{2}-6\sqrt{3}+3\sqrt{5}}\)

\(=\frac{2(3\sqrt{2}-2\sqrt{3}+\sqrt{5})}{3(3\sqrt{2}-2\sqrt{3}+\sqrt{5})}=\frac{2}{3}\)