\(\frac{2}{5\times8}+\frac{2}{8\times11}+\frac{2}{11\times14}+...+\frac{2}{95\times98}\)

\(=\left(\frac{3}{5\times8}+\frac{3}{8\times11}+\frac{3}{11\times14}+...+\frac{3}{95\times98}\right)\times\frac{2}{3}\)

\(=\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{14}+...+\frac{1}{95}-\frac{1}{98}\right)\times\frac{2}{3}\)

\(=\left(\frac{1}{5}-\frac{1}{98}\right)\times\frac{2}{3}\)

\(=\frac{93}{490}\times\frac{2}{3}\)

\(=\frac{93\times2}{490\times3}\)

\(=\frac{31\times1}{245\times1}\)

\(=\frac{31}{245}\)

2/5x8+2/8x11+2/11x14+...+2/95x98

=2(1/5x8+1/8x11+1/11x14+...+1/95x98)       (khoang cach tu 5-8;8-11;11-14;...;95-98 la 3) suy ra =2/3(1/5-1/8+1/8-1/11+1/11-1/14+...+1/95-1/98)

=2/3(1/5-1/98)=2/3x93/5x98=31/245

31/ 245 chính là Đ/S

31/245

\(\dfrac{x}{2\times5}+\dfrac{x}{5\times8}+\dfrac{x}{8\times11}+\dfrac{x}{11\times14}+...+\dfrac{x}{32\times35}=\dfrac{33}{70}\)

\(\dfrac{x}{3}\cdot\left(\dfrac{3}{2\times5}+\dfrac{3}{5\times8}+\dfrac{3}{8\times11}+\dfrac{3}{11\times14}+...+\dfrac{3}{32\times35}\right)=\dfrac{33}{70}\)

\(\dfrac{x}{3}\cdot\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{32}-\dfrac{1}{35}\right)=\dfrac{33}{70}\)

\(\dfrac{x}{3}\cdot\left(\dfrac{1}{2}-\dfrac{1}{35}\right)=\dfrac{33}{70}\)

\(\dfrac{x}{3}\cdot\dfrac{33}{70}=\dfrac{33}{70}\)

\(\dfrac{x}{3}=\dfrac{33}{70}:\dfrac{33}{70}\)

\(\dfrac{x}{3}=1\)

\(x=3\)

\(A=\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+...+\dfrac{3}{100\cdot103}\)

\(=\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{100}-\dfrac{1}{103}\)

\(=\dfrac{98}{515}\)

= 5-2/2x5+8-5/5x8+11-8/8x11+14-11/11x14

=(1/2-1/5)+(1/5-1/8)+(1/8-1/11)+(1/11-1/14)

=(1/2+1/5+1/8+1/11)-(1/5+1/8+1/11+1/14)

=1/2-1/14

=3/7

Vậy B=3/7

B = 3/2x5 + 3/5x8 + 3/8x11 + 3/11x14

B = 1/2 - 1/5 + 1/5 - 1/8 + ..... + 1/14 - 1/17

B = 1/2 - 1/17

B = 15/34

Lời giải:

$\frac{1}{5\times 8}+\frac{1}{8\times 11}+\frac{1}{11\times 14}+...+\frac{1}{m\times (m+3)}=\frac{101}{1540}$

$\frac{8-5}{5\times 8}+\frac{11-8}{8\times 11}+\frac{14-11}{11\times 14}+...+\frac{(m+3)-m}{m\times (m+3)}=\frac{303}{1540}$

$\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{m}-\frac{1}{m+3}=\frac{303}{1540}$

$\frac{1}{5}-\frac{1}{m+3}=\frac{303}{1540}$

$\frac{1}{m+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}$

$\Rightarrow m+3=308$

$\Rightarrow m=308-3=305$

Lời giải:

$\frac{1}{5\times 8}+\frac{1}{8\times 11}+\frac{1}{11\times 14}+...+\frac{1}{m\times (m+3)}=\frac{101}{1540}$

$\frac{8-5}{5\times 8}+\frac{11-8}{8\times 11}+\frac{14-11}{11\times 14}+...+\frac{(m+3)-m}{m\times (m+3)}=\frac{303}{1540}$

$\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{m}-\frac{1}{m+3}=\frac{303}{1540}$

$\frac{1}{5}-\frac{1}{m+3}=\frac{303}{1540}$

$\frac{1}{m+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}$

$\Rightarrow m+3=308$

$\Rightarrow m=308-3=305$

A=3/5x8+3/8x11+3/11x14+...+3/2009x2012+3/2012x2015

A=1/5-1/8+1/8-1/11+1/11-1/14+...+1/2009-1/2012+1/2012-1/2015

A=1/5-1/2015

A=403/2015-1/2015

A=402/2015

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