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$\frac{2}{5\times 8}+\frac{2}{8\times 11}+\frac{2}{11\times 14}+...+\frac{2}{95\times 98}$
$=\left(\frac{3}{5\times 8}+\frac{3}{8\times 11}+\frac{3}{11\times 14}+...+\frac{3}{95\times 98}\right)\times \frac{2}{3}$
$=\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{14}+...+\frac{1}{95}-\frac{1}{98}\right)\times \frac{2}{3}$
$=\left(\frac{1}{5}-\frac{1}{98}\right)\times \frac{2}{3}$
$=\frac{93}{490}\times \frac{2}{3}$
$=\frac{93\times 2}{490\times 3}$
$=\frac{31\times 1}{245\times 1}$
$=\frac{31}{245}$
\(\frac{2}{5\times8}+\frac{2}{8\times11}+\frac{2}{11\times14}+...+\frac{2}{95\times98}\)
\(=\left(\frac{3}{5\times8}+\frac{3}{8\times11}+\frac{3}{11\times14}+...+\frac{3}{95\times98}\right)\times\frac{2}{3}\)
\(=\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{14}+...+\frac{1}{95}-\frac{1}{98}\right)\times\frac{2}{3}\)
\(=\left(\frac{1}{5}-\frac{1}{98}\right)\times\frac{2}{3}\)
\(=\frac{93}{490}\times\frac{2}{3}\)
\(=\frac{93\times2}{490\times3}\)
\(=\frac{31\times1}{245\times1}\)
\(=\frac{31}{245}\)
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
\(\dfrac{x}{2\times5}+\dfrac{x}{5\times8}+\dfrac{x}{8\times11}+\dfrac{x}{11\times14}+...+\dfrac{x}{32\times35}=\dfrac{33}{70}\)
\(\dfrac{x}{3}\cdot\left(\dfrac{3}{2\times5}+\dfrac{3}{5\times8}+\dfrac{3}{8\times11}+\dfrac{3}{11\times14}+...+\dfrac{3}{32\times35}\right)=\dfrac{33}{70}\)
\(\dfrac{x}{3}\cdot\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{32}-\dfrac{1}{35}\right)=\dfrac{33}{70}\)
\(\dfrac{x}{3}\cdot\left(\dfrac{1}{2}-\dfrac{1}{35}\right)=\dfrac{33}{70}\)
\(\dfrac{x}{3}\cdot\dfrac{33}{70}=\dfrac{33}{70}\)
\(\dfrac{x}{3}=\dfrac{33}{70}:\dfrac{33}{70}\)
\(\dfrac{x}{3}=1\)
\(x=3\)
\(\Leftrightarrow x\cdot\dfrac{1}{3}\cdot\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{32\cdot35}\right)=\dfrac{33}{70}\)
=>\(x\cdot\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{32}-\dfrac{1}{35}\right)=\dfrac{33}{70}\)
=>\(x\cdot\dfrac{1}{3}\cdot\dfrac{33}{70}=\dfrac{33}{70}\)
=>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
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:
$\frac{1}{5\times 8}+\frac{1}{8\times 11}+\frac{1}{11\times 14}+...+\frac{1}{m\times (m+3)}=\frac{101}{1540}$
$\frac{8-5}{5\times 8}+\frac{11-8}{8\times 11}+\frac{14-11}{11\times 14}+...+\frac{(m+3)-m}{m\times (m+3)}=\frac{303}{1540}$
$\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{m}-\frac{1}{m+3}=\frac{303}{1540}$
$\frac{1}{5}-\frac{1}{m+3}=\frac{303}{1540}$
$\frac{1}{m+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}$
$\Rightarrow m+3=308$
$\Rightarrow m=308-3=305$
31/ 245 chính là Đ/S
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