Question

so sánh

a) \(\sqrt{10}\) + \(\sqrt{5}\) +1 với \(\sqrt{35}\)

b)\(\sqrt{1+\sqrt{2+\sqrt{3}}}\) và 2

Answer 1

Lời giải:

a) Ta thấy:

\(\sqrt{10}>\sqrt{9}=3; \sqrt{5}>\sqrt{4}=2\)

\(\Rightarrow \sqrt{10}+\sqrt{5}+1>3+2+1=6(1)\)

Mà \(\sqrt{35}< \sqrt{36}=6(2)\)

Từ \((1);(2)\Rightarrow \sqrt{10}+\sqrt{5}+1>\sqrt{35}\)

b)

\(\sqrt{1+\sqrt{2+\sqrt{3}}}=\sqrt{\frac{\sqrt{2}+\sqrt{4+2\sqrt{3}}}{\sqrt{2}}}=\sqrt{\frac{\sqrt{2}+\sqrt{3+1+2\sqrt{3}}}{\sqrt{2}}}\)

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

Mà:

\(\frac{\sqrt{2}+\sqrt{3}+1}{\sqrt{2}}=\frac{2+2\sqrt{3}+\sqrt{2}}{2}< \frac{2+2.\sqrt{4}+2}{2}=4\)

\(\Rightarrow \sqrt{1+\sqrt{2+\sqrt{3}}}< \sqrt{4}=2\)