2/5x8 + 2/8x11 + 2/11x14 + .... + 2/95 x 98
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$\frac{2}{5\times 8}+\frac{2}{8\times 11}+\frac{2}{11\times 14}+...+\frac{2}{95\times 98}$
$=\left(\frac{3}{5\times 8}+\frac{3}{8\times 11}+\frac{3}{11\times 14}+...+\frac{3}{95\times 98}\right)\times \frac{2}{3}$
$=\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{14}+...+\frac{1}{95}-\frac{1}{98}\right)\times \frac{2}{3}$
$=\left(\frac{1}{5}-\frac{1}{98}\right)\times \frac{2}{3}$
$=\frac{93}{490}\times \frac{2}{3}$
$=\frac{93\times 2}{490\times 3}$
$=\frac{31\times 1}{245\times 1}$
$=\frac{31}{245}$
\(A=\dfrac{2}{5\times8}+\dfrac{2}{8\times11}+\dfrac{2}{11\times14}+...+\dfrac{2}{95\times98}\)
\(=2\times\dfrac{1}{3}\times\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{95}-\dfrac{1}{98}\right)\)
\(=\dfrac{2}{3}\times\left(\dfrac{1}{5}-\dfrac{1}{98}\right)\)
\(=\dfrac{2}{3}\times\dfrac{93}{490}\)
\(=\dfrac{31}{245}\)
\(\dfrac{2}{5x8}\) + \(\dfrac{2}{8x11}\) + \(\dfrac{2}{11x14}\)+........+\(\dfrac{2}{95x98}\)
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\(\frac{2}{5\times8}+\frac{2}{8\times11}+\frac{2}{11\times14}+...+\frac{2}{95\times98}\)
\(=\left(\frac{3}{5\times8}+\frac{3}{8\times11}+\frac{3}{11\times14}+...+\frac{3}{95\times98}\right)\times\frac{2}{3}\)
\(=\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{14}+...+\frac{1}{95}-\frac{1}{98}\right)\times\frac{2}{3}\)
\(=\left(\frac{1}{5}-\frac{1}{98}\right)\times\frac{2}{3}\)
\(=\frac{93}{490}\times\frac{2}{3}\)
\(=\frac{93\times2}{490\times3}\)
\(=\frac{31\times1}{245\times1}\)
\(=\frac{31}{245}\)
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
\(\dfrac{x}{2\times5}+\dfrac{x}{5\times8}+\dfrac{x}{8\times11}+\dfrac{x}{11\times14}+...+\dfrac{x}{32\times35}=\dfrac{33}{70}\)
\(\dfrac{x}{3}\cdot\left(\dfrac{3}{2\times5}+\dfrac{3}{5\times8}+\dfrac{3}{8\times11}+\dfrac{3}{11\times14}+...+\dfrac{3}{32\times35}\right)=\dfrac{33}{70}\)
\(\dfrac{x}{3}\cdot\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{32}-\dfrac{1}{35}\right)=\dfrac{33}{70}\)
\(\dfrac{x}{3}\cdot\left(\dfrac{1}{2}-\dfrac{1}{35}\right)=\dfrac{33}{70}\)
\(\dfrac{x}{3}\cdot\dfrac{33}{70}=\dfrac{33}{70}\)
\(\dfrac{x}{3}=\dfrac{33}{70}:\dfrac{33}{70}\)
\(\dfrac{x}{3}=1\)
\(x=3\)
\(3\times\left(\frac{1}{5\times8}+\frac{1}{8\times11}+....+\frac{1}{97\times100}+x\right)=\frac{319}{100}\)
\(\Rightarrow\left(\frac{3}{5\times8}+\frac{3}{8\times11}+\frac{3}{11\times14}+...+\frac{3}{97\times100}\right)+3\times x=\frac{319}{100}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{97}-\frac{1}{100}+3\times x=\frac{319}{100}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{100}+3\times x=\frac{319}{100}\)
\(\Rightarrow\frac{19}{100}+3\times x=\frac{319}{100}\)
\(\Rightarrow3\times x=\frac{319}{100}-\frac{19}{100}\)
\(\Rightarrow3\times x=3\)
\(\Rightarrow x=3:3\)
\(\Rightarrow x=1\)
Vậy x = 1
\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{27}{480}\)
\(\Rightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{27}{480}.\frac{1}{3}\)
\(\Rightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{3}{160}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{1}{160}\)
\(\Rightarrow\frac{1}{x+3}=\frac{31}{160}\)
\(\Rightarrow160=31x+93\)
\(\Rightarrow31x=67\)
\(\Rightarrow x=\frac{67}{31}\)
A=1/5-1/8+1/8-1/11+...+1/602-1/605
=1/5-1/605
=24/121
31/ 245 chính là Đ/S
31/245