Ta co:

1/2+1/6+1/12+1/20+1/30+1/42

=1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7

=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7

=1-1/7

=6/7

Ta co: 6/7=6.10101/7.10101=60606/70707 

Vi 60606/70707=60606/70707 nen 6/7=60606/70707

Vay 1/2+1/6+1/12+1/20+1/30+1/42 = 60606/70707

ta có:

\(\frac{60606}{70707}=\frac{6}{7}\)

ta có:

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}=\)

\(1-\frac{1}{2}+\frac{1}{2}-...-\frac{1}{7}=\frac{6}{7}\)

suy ra 6/7=6/7

suy ra \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}=\frac{60606}{70707}\)

B<A

1 nâng lên lũy thừa nào cũng bằng 1, do đó:

\(1^6=1^{12}=1^{20}=1^{30}=1^{40}=1^3\)

\(\Rightarrow A=1+1+1+1+1=5;B=1\)

\(5>1\Rightarrow A>B\)

\(E=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{2}{2256}\)
\(=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{47.48}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{47}-\dfrac{1}{48}\)
\(=\dfrac{1}{2}-\dfrac{1}{48}\)
\(=\dfrac{23}{48}\)

`A=(20^10+1)/(20^11+1)`

`=>20A=(20^11+20)/(20^11+1)=1+19/(20^11+1)`

Hoàn toàn tương tự: `20B=1+19/(20^12+1)`

Vì `19/(20^12+1)<19/(20^11+1)`

`=>20B<20A`

`=>B<A`