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A = 1/2 - 1/3 - 1/4 + 1/3 - 1/4 - 1/5 + 1/4 - 1/5 - 1/6
A = 1/2 - 1/6
A = 1/3
k mk nha. ths bn nhìu nha
A = 1/2.3 - 1/ 3.4 + 1/3.4 - 1/4.5 + 1/4.5 -1/5.6
= 1/2.3 - 1/5.6
= 1/6 - 1/30
= 2/15
Vậy A = 1/15
Tính:
S = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+\frac{1}{4.5.6}+...+\frac{1}{98.99.100}\)
\(2S=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(2S=\frac{1}{2}-\frac{1}{9900}\)
\(2S=\frac{4949}{9900}\)
\(S=\frac{4949}{19800}\)
Ta xét : \(\frac{1}{1.2}-\frac{1}{2.3}=\frac{2}{1.2.3}\)
\(\frac{1}{2.3}-\frac{1}{3.4}=\frac{2}{2.3.4}\)
...
\(\frac{1}{98.99}-\frac{1}{99.100}=\frac{2}{98.99.100}\)
Ta có : 2S = \(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\)
=> 2S = \(\frac{1}{1.2}-\frac{1}{99.100}\)
=> 2S = \(\frac{4949}{9900}\)
=> S = \(\frac{4949}{19800}\)
1/2x3x4 + 1/3x4x5 + 1/4x5x6 + 1/5x6x7 + ..... + 1/8x9x10
= { 2/2x3x4 + 2/3x4x5 + 2/4x5x6 + .... + 2/8x9x10 } : 2
= { 4-2/2x3x4 + 5-3/3x4x5 + 6-4/4x5x6 + .... + 10-8/8x9x10 } : 2
= { 4/2x3x4 - 2/2x3x4 + 5/3x4x5 - 3/3x4x5 + ... + 10/8x9x10 - 8/8x9x10 } : 2
= { 1/2x3 - 1/3x4 + 1/3x4 - 1/4x5 + ... + 1/8x9 - 1/9x10 } : 2
= { 1/2x3 - 1/9x10 } :2
= { 1/6 - 1/90 } : 2
= 14/90 : 2
= 7/90
\(\frac{3x}{5}=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{6.7.8}\)
Ta có: \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{6.7.8}\)
\(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{6.7.8}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{6.7}-\frac{1}{7.8}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{7.8}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{56}\right)\)
\(=\frac{1}{2}.\frac{27}{56}=\frac{27}{112}\)
\(\frac{3x}{5}=\frac{27}{112}\)
\(\Rightarrow3x=\frac{27.5}{112}\)
\(\Rightarrow3x=\frac{135}{112}\)
\(\Rightarrow x=\frac{45}{112}\)
~Học tốt~
a)\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+....+\(\frac{1}{100.101}\)=1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+....+\(\frac{1}{100}\)-\(\frac{1}{101}\)=1-\(\frac{1}{101}\)=\(\frac{100}{101}\)
b)\(\frac{1}{1.2.3}\)+\(\frac{1}{2.3.4}\)+....+\(\frac{1}{28.29.30}\)=\(\frac{868}{3480}\)=\(\frac{217}{870}\)
c)\(\frac{1}{1.2.3.4}\)+\(\frac{1}{2.3.4.5}\)+....+\(\frac{1}{27.28.29.30}\)=\(\frac{24354}{438480}\)=\(\frac{451}{8120}\)
Đặt A=\(\frac{-1}{2\cdot3\cdot4}\)+\(\frac{-1}{3\cdot4\cdot5}\)+...+\(\frac{-1}{12\cdot13\cdot14}\)
A=(-1)*(\(\frac{1}{2\cdot3\cdot4}\)+\(\frac{1}{3\cdot4\cdot5}\)+...+\(\frac{1}{12\cdot13\cdot14}\))
A=\(\frac{-1}{2}\)*(\(\frac{2}{2\cdot3\cdot4}\)+\(\frac{2}{3\cdot4\cdot5}\)+...+\(\frac{2}{12\cdot13\cdot14}\))
A=\(\frac{-1}{2}\)*(\(\frac{1}{2\cdot3}\)-\(\frac{1}{3\cdot4}\)+\(\frac{1}{3\cdot4}\)-\(\frac{1}{4\cdot5}\)+...+\(\frac{1}{12\cdot13}\)-\(\frac{1}{13\cdot14}\))
A=\(\frac{-1}{2}\)*(\(\frac{1}{6}\)-\(\frac{1}{182}\))
A=\(\frac{-1}{2}\)*\(\frac{44}{273}\)
A=\(\frac{-22}{273}\)
\(\frac{-1}{2.3.4}+\frac{-1}{3.4.5}+\frac{-1}{4.5.6}+...+\frac{-1}{11.12.13}+\frac{-1}{12.13.14}\)
\(=-\frac{1}{2}.\left(\frac{2}{2.3.4}+\frac{2}{3.4.5}+\frac{2}{4.5.6}+...+\frac{2}{11.12.13}+\frac{2}{12.13.14}\right)\)
\(=-\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+\frac{1}{4.5}-\frac{1}{5.6}+...+\frac{1}{11.12}-\frac{1}{12.13}+\frac{1}{12.13}-\frac{1}{13.14}\right)\)
\(=-\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{13.14}\right)=-\frac{1}{2}.\frac{44}{273}=-\frac{22}{273}\)
C = \(\frac{3}{2.3.4}+\frac{3}{3.4.5}+...+\frac{3}{98.99.100}\)
C = \(3.\frac{1}{2}.\left(\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\right)\)
C = \(\frac{3}{2}.\left(\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{100-98}{98.99.100}\right)\)
C = \(\frac{3}{2}.\left(\frac{4}{2.3.4}-\frac{2}{2.3.4}+\frac{5}{3.4.5}-\frac{3}{3.4.5}+...+\frac{100}{98.99.100}-\frac{98}{98.99.100}\right)\)
C = \(\frac{3}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
C = \(\frac{3}{2}.\left(\frac{1}{2.3}-\frac{1}{99.100}\right)\)
C = \(\frac{3}{2}.\frac{1649}{9900}\)
C = \(\frac{1649}{6600}\)
