Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\Leftrightarrow x\cdot\dfrac{1}{3}\cdot\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{32\cdot35}\right)=\dfrac{33}{70}\)
=>\(x\cdot\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{32}-\dfrac{1}{35}\right)=\dfrac{33}{70}\)
=>\(x\cdot\dfrac{1}{3}\cdot\dfrac{33}{70}=\dfrac{33}{70}\)
=>x=3
\(A=\dfrac{3}{5\times8}+\dfrac{3}{8\times11}+...+\dfrac{3}{2009\times2012}\)
\(A=\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{2009}-\dfrac{1}{2012}\)
\(=\dfrac{1}{5}-\dfrac{1}{2012}=\dfrac{2007}{10060}\)
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
= 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
\(A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{11}-\frac{1}{14}\)
\(A=\frac{1}{2}-\frac{1}{14}\)
\(A=\frac{3}{7}\)
\(A=\)\(\frac{3}{5\times2}+\frac{3}{5\times8}+\frac{3}{8\times11}+...+\frac{3}{11\times14}\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{11}-\frac{1}{14}\)
\(=\frac{1}{2}-\frac{1}{14}\)
\(=\frac{3}{7}\)
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:
$\frac{1}{5\times 8}+\frac{1}{8\times 11}+\frac{1}{11\times 14}+...+\frac{1}{m\times (m+3)}=\frac{101}{1540}$
$\frac{8-5}{5\times 8}+\frac{11-8}{8\times 11}+\frac{14-11}{11\times 14}+...+\frac{(m+3)-m}{m\times (m+3)}=\frac{303}{1540}$
$\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{m}-\frac{1}{m+3}=\frac{303}{1540}$
$\frac{1}{5}-\frac{1}{m+3}=\frac{303}{1540}$
$\frac{1}{m+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}$
$\Rightarrow m+3=308$
$\Rightarrow m=308-3=305$
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\)
\(=\frac{1}{2}-\frac{1}{14}\)
\(=\frac{7}{14}-\frac{1}{14}\)
\(=\frac{6}{14}\)
\(=\frac{3}{7}\)
3/2x5 + 3/5x8 + 3/8x11 + 3/11x14
= 3/2 - 3/5 + 3/5 - 3/8 + 3/8 - 3/11 + 3/11 - 3/14
= 3/2 - 3/14
= 21/14 - 3/14
= 18/14
= 9/5
\(A=\dfrac{1}{1\times4}+\dfrac{1}{4\times7}+\dfrac{1}{7\times10}+...+\dfrac{1}{100\times103}\)
\(=\dfrac{1}{3}\left(\dfrac{3}{1\times4}+\dfrac{3}{4\times7}+...+\dfrac{3}{100\times103}\right)\)
\(=\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)
\(=\dfrac{1}{3}\left(1-\dfrac{1}{103}\right)=\dfrac{1}{3}.\dfrac{102}{103}=\dfrac{34}{103}\)
\(A=\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+...+\dfrac{3}{100\cdot103}\)
\(=\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{100}-\dfrac{1}{103}\)
\(=\dfrac{98}{515}\)