\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}\\ =1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\\ =1-\dfrac{1}{11}=\dfrac{10}{11}\)

\(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}\\ =1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\\ =1-\dfrac{1}{11}\\ =\dfrac{10}{11}\)

= \(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\)

\(=1-\dfrac{1}{11}\)

\(=\dfrac{10}{11}\)

\(C=\dfrac{-5}{7}+\dfrac{-2}{7}+\dfrac{3}{4}+\dfrac{1}{4}+\dfrac{-1}{5}=-1+1-\dfrac{1}{5}=\dfrac{-1}{5}\)

\(\dfrac{5}{1\cdot4}+\dfrac{5}{4\cdot7}+\dfrac{5}{7\cdot11}+...+\dfrac{5}{\left(3x+1\right)\cdot\left(3x+4\right)}\\ =\dfrac{5}{3}\cdot\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot11}+...+\dfrac{3}{\left(3x+1\right)\cdot\left(3x+4\right)}\right)\\ =\dfrac{5}{3}\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{3x+1}-\dfrac{1}{3x+4}\right)\\ =\dfrac{5}{3}\cdot\left(1-\dfrac{1}{3x+4}\right)\\ =\dfrac{5}{3}-\dfrac{5}{9x+12}\)

1100444-88888=

a)\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)

\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)

\(=\frac{1}{2}.\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+\frac{11-9}{9.11}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{11}\right)\)

\(\frac{10}{22}\)

thế phần b và c đâu???

\(\dfrac{-1}{12},\dfrac{-3}{4},\dfrac{2}{9},\dfrac{7}{6}\)

\(-\dfrac{1}{12},-\dfrac{3}{4},\dfrac{2}{9},\dfrac{7}{6}\)

`1/5 . 4/7 + 3/7 . 1/5 -1/5`

`=1/5 . 4/7 + 3/7 . 1/5 -1/5 . 1`

`=1/5 . ( 4/7+3/7-1)`

`=1/5 . ( 7/7-1)`

`= 1/5 . 0`

`=0`

\(\dfrac{1}{5}\times\dfrac{4}{7}+\dfrac{3}{7}\times\dfrac{1}{5}-\dfrac{1}{5}=\dfrac{1}{5}\times\left(\dfrac{4}{7}+\dfrac{3}{7}-1\right)=\dfrac{1}{5}\times0=0\)

= \(\dfrac{1}{5}\).(\(\dfrac{4}{7}\)+\(\dfrac{3}{7}\)+1)

= \(\dfrac{1}{5}\)+ 0

\(\dfrac{1}{5}\)

Lời giải:
$S=\frac{1}{7^2}+\frac{2}{7^3}+\frac{3}{7^4}+...+\frac{69}{7^{70}}$

$7S=\frac{1}{7}+\frac{2}{7^2}+\frac{3}{7^3}+...+\frac{69}{7^{69}}$

$6S=7S-S=\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+....+\frac{1}{7^{69}}-\frac{69}{7^{70}}$

$42S=1+\frac{1}{7}+\frac{1}{7^2}+...+\frac{1}{7^{68}}-\frac{69}{7^{69}}$

$\Rightarrow 42S-6S=(1+\frac{1}{7}+\frac{1}{7^2}+...+\frac{1}{7^{68}}-\frac{69}{7^{69}})-(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+....+\frac{1}{7^{69}}-\frac{69}{7^{70}})$

$\Rightarrow 36S=1-\frac{69}{7^{69}}-\frac{1}{7^{69}}+\frac{69}{7^{70}}$

Hay $36S=1-\frac{69.7-7-69}{7^{70}}=1-\frac{407}{7^{70}}$

$\Rightarrow S=\frac{1}{36}(1-\frac{407}{7^{70}})$

A= 1/3 + 1/3^2 + ... + 1/3^8

3A= 3. (1/3+ 1/3^2+ ... + 1/3^8)

3A=1+ 1/3 + 1/3^2+ ... +1/3^7

=> 3A - A= (1 + 1/3 + 1/3^2 + ... + 1/3^7) - (1/3 + 1/3^2+ ... + 1/3^8)

=> 2A= 1 - 1/ 3^8

2A= 6560/6561

A= 6560/6561 : 2

A= 3280/6561