Ta có: 
1/4 = 1/3( 1-1/4) 
1/28 = 1/3( 1/4 - 1/7) 
1/70 = 1/3( 1/7 - 1/10) 
.............................. 
1/10300 = 1/3( 1/100 - 1/103) 
Cộng vế với vế ta có: 
S = 1/4+1/28+1/70+1/130+...+1/10300 = 1/3( 1-1/103) 
S = 34/103

\(3M=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\)

\(3M=\frac{4-1}{1.4}+\frac{7-4}{4.7}+...+\frac{100-97}{97.100}\)

\(3M=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\)

\(3M=1-\frac{1}{100}\)

\(3M=\frac{99}{100}\)

\(M=\frac{33}{100}\)

$\dfrac14+\dfrac1{28}+\dfrac1{70}+\cdots+\dfrac1{9700}$

$=\dfrac1{2\cdot2}+\dfrac1{4\cdot7}+\dfrac1{7\cdot10}+\cdots+\dfrac1{97\cdot100}$

$=\dfrac13\left(\dfrac12-\dfrac15\right)+\dfrac13\left(\dfrac14-\dfrac17\right)+\dfrac13\left(\dfrac17-\dfrac1{10}\right)+\cdots+\dfrac13\left(\dfrac1{97}-\dfrac1{100}\right)$

$=\dfrac13\left(\dfrac12-\dfrac15+\dfrac14-\dfrac17+\dfrac17-\dfrac1{10}+\cdots+\dfrac1{97}-\dfrac1{100}\right)$

$=\dfrac13\left(\dfrac12+\dfrac14-\dfrac15-\dfrac1{100}\right)$

$=\dfrac13\left(\dfrac{50+25-20-1}{100}\right)$

$=\dfrac13\cdot\dfrac{54}{100}$

$=\dfrac{9}{50}$

Theo đề bài:

$\dfrac{9}{50}=\dfrac{0,33x}{2009}$

$\Leftrightarrow\dfrac{9}{50}=\dfrac{33x}{100\cdot2009}$

$\Leftrightarrow 9\cdot100\cdot2009=50\cdot33x$

$\Leftrightarrow x=\dfrac{9\cdot100\cdot2009}{50\cdot33}$

$\Leftrightarrow x=\dfrac{6\cdot2009}{11}$

$\Leftrightarrowx=\dfrac{12054}{11}$.

M=1/4+1/4.7+1/7.10+1/10.13+.............+1/88.91

3M=3/4+3/4.7+3/7.10+3/10.13+........+3/88.91

3M= 3/4+1/4-1/7+1/7-1/10+1/10-1/13+......+1/88-1/91

3M=3/4+1/4-1/91=1-1/91=90/91 

----->M= 30/91

$M=\dfrac14+\dfrac1{28}+\dfrac1{70}+\dfrac1{130}+\cdots+\dfrac1{8008}$

$=\dfrac1{2\cdot2}+\dfrac1{4\cdot7}+\dfrac1{7\cdot10}+\dfrac1{10\cdot13}+\cdots+\dfrac1{88\cdot91}$

$=\dfrac14+\dfrac13\left(\dfrac14-\dfrac17\right)+\dfrac13\left(\dfrac17-\dfrac1{10}\right)+\dfrac13\left(\dfrac1{10}-\dfrac1{13}\right)+\cdots+\dfrac13\left(\dfrac1{88}-\dfrac1{91}\right)$

$=\dfrac14+\dfrac13\left(\dfrac14-\dfrac17+\dfrac17-\dfrac1{10}+\dfrac1{10}-\dfrac1{13}+\cdots+\dfrac1{88}-\dfrac1{91}\right)$

$=\dfrac14+\dfrac13\left(\dfrac14-\dfrac1{91}\right)$

$=\dfrac14+\dfrac{29}{364}$

$=\dfrac{91+29}{364}$

$=\dfrac{120}{364}$

$=\dfrac{30}{91}$.

\(4-\dfrac{1}{4}-\dfrac{1}{28}-\dfrac{1}{70}-.....-\dfrac{1}{2002.2005}\)

\(=4-\left(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+.....+\dfrac{1}{2002.2005}\right)\)

\(=4-\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+.....+\dfrac{1}{2002}-\dfrac{1}{2005}\right)\)\(=4-\left(1-\dfrac{1}{2005}\right)\)

\(=4-1+\dfrac{1}{2005}=3+\dfrac{1}{2005}=\dfrac{6016}{2005}\)

\(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+\frac{1}{130}+\frac{1}{208}\)

\(=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{13.16}\)

\(=\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}\right)\)

\(=\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}\right)\)

\(=\frac{1}{3}.\left(1-\frac{1}{16}\right)\)

\(=\frac{1}{3}.\frac{15}{16}=\frac{5}{16}\)

1/1.4+1/4.7+1/7.10+..+1/13.16

1-1/4+1/4-1/7+..+1/13-1/16

1-1/16

15/16

Vào câu hỏi tương tự nha bạn !

$M=\dfrac14+\dfrac1{28}+\dfrac1{70}+\dfrac1{130}+\cdots$ (có $30$ số hạng)

$=\dfrac1{2\cdot2}+\dfrac1{4\cdot7}+\dfrac1{7\cdot10}+\dfrac1{10\cdot13}+\cdots+\dfrac1{88\cdot91}$

Với mọi số hạng từ số hạng thứ hai:

$\dfrac1{a(a+3)}=\dfrac13\left(\dfrac1a-\dfrac1{a+3}\right)$

Do đó:

$M=\dfrac14+\dfrac13\left(\dfrac14-\dfrac17\right)+\dfrac13\left(\dfrac17-\dfrac1{10}\right)+\cdots+\dfrac13\left(\dfrac1{88}-\dfrac1{91}\right)$

$=\dfrac14+\dfrac13\left(\dfrac14-\dfrac1{91}\right)$

$=\dfrac14+\dfrac{29}{364}$

$=\dfrac{91+29}{364}$

$=\dfrac{120}{364}$

$=\dfrac{30}{91}$.

A = 1/4 + 1/28 + 1/70 +...+ 1/9700

A = 1/1.4 + 1/4.7 + 1/7.10 +...+ 1/97.100

3A = 3/1.4 + 3/4.7 + 3/7.10 +...+ 3/97.100

3A = 1 - 1/100

3A = 99/100

A=99/100:3=33/100

\(=\frac{1}{1.4}+\frac{1}{4.7}+..+\frac{1}{97.100}\)

\(=\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{97.100}\right)\)

\(=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{100}\right)\)

\(=\frac{1}{3}.\frac{99}{100}=\frac{33}{100}\)