\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+.....+\frac{1}{\sqrt{100}}>\frac{1}{\sqrt{100}}+\frac{1}{\sqrt{100}}+....+\frac{1}{\sqrt{100}}\)

\(\Leftrightarrow\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+....+\frac{1}{\sqrt{100}}>100.\frac{1}{\sqrt{100}}=10.\)

1)  \(A^2=2+2.\frac{\sqrt{\left(8+\sqrt{15}\right)\left(8-\sqrt{15}\right)}}{2}\)

              \(2+\sqrt{64-15}=2+\sqrt{49}=2+7=9\) mà A>0

=> A=3

2) \(A=\sqrt{4-\sqrt{15}}\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right).\)

 \(A=\sqrt{\left(4-\sqrt{15}\right)\left(4+\sqrt{15}\right)}\sqrt{4+\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right).\)

​​\(A=\sqrt{4+\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right).\)

\(A^2=\left(4+\sqrt{15}\right)\left(16-4\sqrt{15}\right)\)

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

Mà A >0 

=> A=2

Mà 4>3

=> \(\sqrt{4}=2>\sqrt{3}\)

=> \(A>\sqrt{3}\)

\(\sqrt{\frac{10}{17}}va\frac{3}{4}\)

Ta có \(\frac{10}{17}>\frac{9}{16}\)

\(\Rightarrow\sqrt{\frac{10}{17}}>\sqrt{\frac{9}{16}}\)

\(\Rightarrow\sqrt{\frac{10}{17}}>\frac{3}{4}\)

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