\(A< \frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}=\frac{5}{5}=1=B\)

a/

\(\frac{2001}{2004}=\frac{2004-3}{2004}=1-\frac{3}{2004}=1-\frac{1}{668}.\)

\(\frac{39}{40}=\frac{40-1}{40}=1-\frac{1}{40}\)

Ta có \(40< 668\Rightarrow\frac{1}{40}>\frac{1}{668}\Rightarrow1-\frac{1}{40}< 1-\frac{1}{668}\Rightarrow\frac{39}{40}< \frac{2001}{2004}\)

b/

\(A< \frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}=1=B\)

\(M=\frac{1}{1.2}+\frac{2}{1.2.3}+.....+\frac{9}{1.2.3.....10}\)

\(M=\frac{2-1}{1.2}+\frac{3-1}{1.2.3}+....+\frac{10-1}{1.2......10}\)

\(M=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{6}+....+\frac{10}{1.2.....10}-\frac{1}{1.2.....10}\)

\(M=1-\frac{1}{1.2.3......10}\)

\(M=1-\frac{1}{3628800}\)

Vì \(1=1\)mà \(\frac{1}{3628800}< 1\)nên \(1-\frac{1}{3628800}< 1\)

Vậy \(M< 1\)

       \(M=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2016}}\)

\(\Rightarrow\)\(2M=2+1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{2015}}\)

\(\Rightarrow\)\(2M-M=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\right)\)

\(\Rightarrow\)\(M=2-\frac{1}{2^{2016}}< 2\)

Vậy  M < 2 

\(M=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{10^2}\)

\(>1+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\)

\(=1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\)

\(=1+\frac{1}{2}-\frac{1}{11}\)

\(>1+\frac{1}{2}-\frac{1}{6}=\frac{4}{3}\)

ta có \(A=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right).......\left(\frac{1}{10}-1\right)\)

\(A=-\left(\frac{1}{2}.\frac{2}{3}.....\frac{9}{10}\right)\)

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

vi\(-\frac{1}{10}>-\frac{1}{9}\)

do đó A>\(\frac{-1}{9}\)

M = 1 + 1/2 + 1/2² + ... + 1/2²⁰¹⁵ + 1/2²⁰¹⁶

=> 2.M = 2 + 1 + 1/2 + ... + 1/2²⁰¹⁴ + 1/2²⁰¹⁵

=> 2.M = 3 + 1/2 +...+ 1/2²⁰¹⁴ + 1/2²⁰¹⁵

=> 2.M - M = 3 -1 + 1/2²⁰¹⁶

=> M = 2 + 1/2²⁰¹⁶ 

=> M > 2 ( Do 2 + 1/2²⁰¹⁶  > 2 )

cam on ban

\(\frac{1}{15}+\frac{1}{16}+\frac{1}{17}+...+\frac{1}{30}>\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}=\frac{16}{30}>\frac{15}{30}=\frac{1}{2}\)