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Ta có: \(M=\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+....+\frac{1}{\sqrt{224}+\sqrt{225}}\)
\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{225}-\sqrt{224}\)
\(=-1+\sqrt{225}=-1+15=14\)
Và \(N=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{63}}\)
\(=14,47706...>14=M\)
\(\frac{1}{\sqrt{2}}=\frac{2}{2\sqrt{2}}< \frac{2}{\sqrt{2}+\sqrt{1}}=\frac{2\left(\sqrt{2}-1\right)}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}=2\left(\sqrt{2}-1\right)\)
\(\frac{1}{\sqrt{3}}=\frac{2}{2\sqrt{3}}< \frac{2}{\sqrt{3}+\sqrt{2}}=\frac{2\left(\sqrt{3}-\sqrt{2}\right)}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}=2\left(\sqrt{3}-\sqrt{2}\right)\)
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\(\frac{1}{\sqrt{225}}=\frac{2}{2\sqrt{225}}< \frac{2}{\sqrt{225}+\sqrt{224}}=\frac{2\left(\sqrt{225}-\sqrt{224}\right)}{\left(\sqrt{225}+\sqrt{224}\right)\left(\sqrt{225}-\sqrt{224}\right)}\)\(=2\left(\sqrt{225}-\sqrt{224}\right)\)
\(\Rightarrow\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{225}}< 2\left(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{225}-\sqrt{224}\right)\)
\(\Rightarrow\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{225}}< 2\left(\sqrt{225}-1\right)=2\left(15-1\right)=28\)
vế phải < \(2.\left(\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{3}}+...+\frac{1}{2\sqrt{225}}\right)\)
<\(2\left(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{224}+\sqrt{225}}\right)\)
\( =2.\left(-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-...-\sqrt{224}+\sqrt{225}\right)\)
=\(2.\left(-1+\sqrt{225}\right)=2.14=28\)
