tất cả các tỉ số bạn nêu đều bằng 2/3

\(A=\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+\frac{1}{64}+\frac{1}{100}+\frac{1}{144}+\frac{1}{196}\)

\(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}\)

\(A=\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{7^2}\right)\)

\(< \frac{1}{2^2}\left(1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{6}-\frac{1}{7}\right)\)

\(=\frac{1}{2^2}\left(1-\frac{1}{7}\right)\)

\(=\frac{1}{2^2}\cdot\frac{6}{7}\)

\(=\frac{3}{14}\)

\(< \frac{1}{2}\)

Ta có: \(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{72}+\frac{1}{90}\)

\(B=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\)

\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)

\(B=1-\frac{1}{10}\)

\(B=\frac{10}{10}-\frac{1}{10}\)

\(B=\frac{9}{10}\)

Vậy: \(B=\frac{9}{10}\)

\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}+\frac{1}{9.10}\)

\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)

\(B=1-\frac{1}{10}\)

\(B=\frac{9}{10}\)

Vì \(\frac{9}{10}< 1\)nên B < 1

Vậy B < 1

Ta có 

\(B=\frac{1}{2}+\frac{1}{6} +...+\frac{1}{90}\)

  \(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\)

\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)

\(B=1-\frac{1}{10}\)

Vì 0<\(\frac{1}{10}\)<1

=>1-\(\frac{1}{10}\)<1

=>B<1

P = 1/3 + 1/16 + 1/19 + 1/21 + 1/61 + 1/72 + 1/83 + 1/94

P = 0, 5490527821

3/5 = 0, 6

Mà 0, 5490527821 < 0, 6

Nên: P < 3/5

\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{36}}\)\(=\frac{2}{\sqrt{1}+\sqrt{1}}+\frac{2}{\sqrt{2}+\sqrt{2}}+\frac{2}{\sqrt{3}+\sqrt{3}}+...+\frac{2}{\sqrt{36}+\sqrt{36}}\)

\(< \frac{2}{\sqrt{1}}+\frac{2}{\sqrt{1}+\sqrt{2}}+\frac{2}{\sqrt{2}+\sqrt{3}}+...+\frac{2}{\sqrt{35}+\sqrt{36}}\)

\(=2\left(\sqrt{1}+\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{36}-\sqrt{35}\right)\)

\(=2.\sqrt{36}=12< 14\)

Vậy \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{36}}< 14\)

Ta có : 1/61 +1/72 +1/83 +1/94 < 1/60 x4 =1/15

1/16 +1/19 +1/21 < 1/15 x3 =3/15

1/3 =5/15

Do đó: B < 1/15 + 3/15 +5/15 = 9/15= 3/5

Vậy B< 3/5

                                          \(\text{Bài giải}\)

\(\text{Ta có : }\)

 \(B=\frac{1}{3}+\frac{1}{16}+\frac{1}{19}+\frac{1}{21}+\frac{1}{61}+\frac{1}{72}+\frac{1}{83}+\frac{1}{94}\)

B = 0,333333333 + 0,0625 + 0,0526315789 + 0,0476190476 + 0,0163934426 + 0,0138888889 + 0,0120481928 + 0,0106382979

B = 0,549052782

a/ \(\sqrt{17}+\sqrt{5}+1>\sqrt{16}+\sqrt{4}+1=4+2+1=7\)

\(\sqrt{45}< \sqrt{49}=7\)

\(\Rightarrow\sqrt{17}+\sqrt{5}+1>\sqrt{45}\)

b/ Ta có:

\(\sqrt{n}< \sqrt{n+1}\)

\(\Rightarrow2\sqrt{n}< \sqrt{n+1}+\sqrt{n}\)

\(\Rightarrow\dfrac{1}{\sqrt{n}}>\dfrac{2}{\sqrt{n+1}+\sqrt{n}}=2\left(\sqrt{n+1}-\sqrt{n}\right)\)

Áp dụng vào bài toán được

\(1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{36}}>2\left(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{37}-\sqrt{36}\right)\)

\(=2\left(\sqrt{37}-1\right)>6\)