Question

[Dương Thị Thu Ngọc]

So sánh:

a)a=\(\sqrt{2+\sqrt{3}}\) và b=\(\dfrac{\sqrt{3}+1}{\sqrt{2}}\)

b)b=\(\sqrt{5-\sqrt{13+4\sqrt{3}}}\) và c=\(\sqrt{3}-1\)

c)\(\sqrt{n+2}+\sqrt{n+1}\) và\(\sqrt{n+1}-\sqrt{n}\) (n∈N*)

Answer 1

[Akai Haruma]

Lời giải:

a)

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

b)

\( b=\sqrt{5-\sqrt{12+1+2\sqrt{12}}}=\sqrt{5-\sqrt{(\sqrt{12}+1)^2}}\)

\(=\sqrt{5-(\sqrt{12}+1)}=\sqrt{4-\sqrt{12}}\)

\(=\sqrt{4-2\sqrt{3}}=\sqrt{3+1-2\sqrt{3}}=\sqrt{(\sqrt{3}-1)^2}=\sqrt{3}-1=c\)

c)

\(\sqrt{n+2}>\sqrt{n+1}; \sqrt{n+1}> -\sqrt{n}\)

\(\Rightarrow \sqrt{n+2}+\sqrt{n+1}> \sqrt{n+1}-\sqrt{n}\)