Question

[Mai Linh]

So sánh:

a, \(\frac{3\sqrt{7}+5\sqrt{2}}{\sqrt{5}}\) và \(\sqrt{35}+\sqrt{10}\)

b, \(\frac{\sqrt{3+\sqrt{5}}}{\sqrt{2}}\) và \(\frac{1+\sqrt{5}}{2}\)

c, \(\frac{2+\sqrt{2}}{2-\sqrt{2}}+\frac{2-\sqrt{2}}{2+\sqrt{2}}\) và \(4\sqrt{2}\)

d, \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\) và \(\sqrt{3}\)

Answer 1

[Nguyễn Việt Lâm]

\(\frac{3\sqrt{7}+5\sqrt{2}}{\sqrt{5}}=\frac{3}{5}\sqrt{35}+\sqrt{10}< \sqrt{35}+\sqrt{10}\)

\(\frac{\sqrt{3+\sqrt{5}}}{\sqrt{2}}=\frac{\sqrt{6+2\sqrt{5}}}{2}=\frac{\sqrt{5}+1}{2}\)

\(\frac{2+\sqrt{2}}{2-\sqrt{2}}+\frac{2-\sqrt{2}}{2+\sqrt{2}}=\frac{\left(2+\sqrt{2}\right)^2+\left(2-\sqrt{2}\right)^2}{2}=\frac{12}{2}=6>4\sqrt{2}\) (do \(36>32\))

\(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}=\frac{\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}}{\sqrt{2}}=\frac{\sqrt{7}+1-\left(\sqrt{7}-1\right)}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}< \sqrt{3}\)