Ta có : 

1/6 < 1/5 , 1/7 < 1/5 , ... 1/19 < 1/5

=> 1/6 + 1/7 + ...+ 1/19 < 1/5 + 1/5 + ...+ 1/5

=> 1/6 + 1/7 + ...+ 1/19  < 1/5 . 14 

=> 1/6 + 1/7 + ...+ 1/19 < 14/5 = 2 , 8 

\(A=\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+...+\frac{1}{20}\)

\(=\left(\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}\right)+\frac{1}{12}+\left(\frac{1}{13}+...+\frac{1}{16}\right)+\left(\frac{1}{17}+...+\frac{1}{20}\right)\)

\(>\left(\frac{1}{9}+\frac{1}{9}+\frac{1}{9}\right)+\left(\frac{1}{12}+\frac{1}{12}+\frac{1}{12}\right)+\frac{1}{12}+\left(\frac{1}{16}+...+\frac{1}{16}\right)+\left(\frac{1}{24}+...+\frac{1}{24}\right)\)

\(=\frac{1}{3}+\frac{1}{4}+\frac{1}{12}+\frac{1}{4}+\frac{1}{6}=1+\frac{1}{12}\)

\(B=\frac{1}{5}+\frac{1}{6}+...+\frac{1}{18}+\frac{1}{19}\)

\(=\left(\frac{1}{5}+...+\frac{1}{9}\right)+\left(\frac{1}{10}+...+\frac{1}{14}\right)+\left(\frac{1}{15}+...+\frac{1}{19}\right)\)

\(< \left(\frac{1}{5}+...+\frac{1}{5}\right)+\left(\frac{1}{10}+...+\frac{1}{10}\right)+\left(\frac{1}{15}+...+\frac{1}{15}\right)\)

\(=\frac{5}{5}+\frac{5}{10}+\frac{5}{15}=1+\frac{5}{6}\)

\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}+\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{20}-\dfrac{1}{20}\right)\)

\(=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{20}\right)\)

\(=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{10}\right)\)

\(=\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{20}\) (đpcm)

Có: 1/5 =1/5

1/6<1/5

1/7<1/5

1/8<1/5

1/9<1/5

=> 1/5+1/6+1/7+1/8+1/9<1/5+1/5+1/5+1/5+1/5=1.

Vậy 1/5+1/6+1/7+1/8+1/9<1(đpcm).

(1/5 + 1/6 + 1/7 + 1/8 + 1/9)<(1/5 x 5)

(Vì 5 số hạng biểu thức đề cho có 4 số hạng nhỏ hơn 1/5 và chỉ có 1/5 = 1/5)

⇒ (1/5 + 1/6 + 1/7 + 1/8 + 1/9) < 1

Vậy...

Do1/5+1/6+1/7+1/8+1/9<1/5+1/5+1/5+1/5+1/5=1

Ta có : \(\dfrac{1}{6}>\dfrac{1}{10}\)

\(\dfrac{1}{7}>\dfrac{1}{10}\)

\(\dfrac{1}{8}>\dfrac{1}{10}\)

\(\dfrac{1}{9}>\dfrac{1}{10}\)

\(\dfrac{1}{10}=\dfrac{1}{10}\)

Cộng tất cả các vế ( phải theo phải ) ( trái theo trái ta được )

\(\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}>\dfrac{1}{10}+\dfrac{1}{10}+\dfrac{1}{10}+\dfrac{1}{10}+\dfrac{1}{10}\)

\(\Rightarrow\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}>\dfrac{5}{10}=\dfrac{1}{2}\)

Ta có:
\(\dfrac{1}{6}>\dfrac{1}{10}\)
\(\dfrac{1}{7}>\dfrac{1}{10}\)
\(\dfrac{1}{8}>\dfrac{1}{10}\)
\(\dfrac{1}{9}>\dfrac{1}{10}\)
\(\dfrac{1}{10}=\dfrac{1}{10}\)
Do đó ta có:
\(\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}>\dfrac{1}{10}+\dfrac{1}{10}+\dfrac{1}{10}+\dfrac{1}{10}+\dfrac{1}{10}\)
\(\Rightarrow\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}>\dfrac{1}{10}\times5\)
\(\Rightarrow\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}>\dfrac{5}{10}=\dfrac{1}{2}\)
Vậy \(\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}>\dfrac{1}{2}\)

8:

\(A=\dfrac{20^{10}-1+2}{20^{10}-1}=1+\dfrac{2}{20^{10}-1}\)

\(B=\dfrac{20^{10}-3+2}{20^{10}-3}=1+\dfrac{2}{20^{10}-3}\)

mà 20^10-1>20^10-3

nên A<B