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Ta có : \(S=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}\)
\(\Rightarrow3S=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}\)
\(\Rightarrow3S=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\)
\(\Rightarrow3S=\frac{1}{2}-\frac{1}{14}=\frac{3}{7}\)
\(\Rightarrow S=\frac{3}{7}.\frac{1}{3}=\frac{1}{7}\)
3S= 1/2 - 1/5 + 1/5 - 1/8 + ... + 1/11 - 1/14
3S= 1/2 - 1/14
S= 3/7 / 3
S= 1/7
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\(\frac{1}{2\times5}+\frac{1}{5\times8}+\frac{1}{8\times11}+\frac{1}{11\times14}+\frac{1}{14\times17}+\frac{1}{17\times20}\)
\(=\frac{1}{3}\times\left(\frac{3}{2\times5}+\frac{3}{5\times8}+\frac{3}{8\times11}+\frac{3}{11\times14}+\frac{3}{14\times17}+\frac{3}{17\times20}\right)\)
\(=\frac{1}{3}\times\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\right)\)
\(=\frac{1}{3}\times\left(\frac{1}{2}-\frac{1}{20}\right)\)
\(=\frac{1}{3}\times\frac{9}{20}\)
\(=\frac{3}{20}\)
_Chúc bạn học tốt_
Đặt \(A=\frac{1}{2x5}+\frac{1}{5x8}+..+\frac{1}{17x20}\)
\(3xA=3x\left(\frac{1}{2x5}+\frac{1}{5x8}+...+\frac{1}{17x20}\right)\)
\(3xA=\frac{3}{2x5}+\frac{3}{5x8}+....+\frac{3}{17x20}\)
\(3xA=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+..+\frac{1}{17}-\frac{1}{20}\)
\(3xA=\frac{1}{2}-\frac{1}{20}\)
\(3xA=\frac{9}{20}\)
\(\Rightarrow A=\frac{3}{20}\)
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\(A=\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+...+\dfrac{3}{100\cdot103}\)
\(=\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{100}-\dfrac{1}{103}\)
\(=\dfrac{98}{515}\)
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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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= 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:
$\frac{1}{5\times 8}+\frac{1}{8\times 11}+\frac{1}{11\times 14}+...+\frac{1}{m\times (m+3)}=\frac{101}{1540}$
$\frac{8-5}{5\times 8}+\frac{11-8}{8\times 11}+\frac{14-11}{11\times 14}+...+\frac{(m+3)-m}{m\times (m+3)}=\frac{303}{1540}$
$\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{m}-\frac{1}{m+3}=\frac{303}{1540}$
$\frac{1}{5}-\frac{1}{m+3}=\frac{303}{1540}$
$\frac{1}{m+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}$
$\Rightarrow m+3=308$
$\Rightarrow m=308-3=305$