B = \(\frac{7}{10\cdot11}\)+ \(\frac{7}{11\cdot12}\)+\(\frac{7}{12\cdot13}\)+...+ \(\frac{7}{69\cdot70}\)
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\(\frac{7}{10\cdot11}+\frac{7}{11\cdot12}+....+\frac{7}{69\cdot70}\)
\(\frac{7}{10.11}+\frac{7}{11.12}+...+\frac{7}{69.70}\)
\(=7.\left(\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{69.70}\right)\)
\(=7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(=7.\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(=7.\frac{3}{35}\)
\(=\frac{3}{5}\)
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\(\frac{7}{10\cdot11}+\frac{7}{11\cdot12}+...+\frac{7}{69\cdot70}\)
\(=7\cdot\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(=7\cdot\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(=7\cdot\frac{3}{35}\)
\(=\frac{3}{5}\)
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\(\frac{7}{10.11}+\frac{7}{11.12}+...\frac{7}{69.70}\\ =\frac{7}{10}-\frac{7}{11}+\frac{7}{11}-\frac{7}{12}+...+\frac{7}{69}-\frac{7}{70}\)
\(=\frac{7}{10}-\frac{7}{70}\)
\(=\frac{3}{5}\)
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\(A=\frac{7}{10\cdot11}+\frac{7}{11\cdot12}+\frac{7}{12\cdot13}+...+\frac{7}{69\cdot70}\)
\(B=\frac{1}{25\cdot27}+\frac{1}{27\cdot29}+\frac{1}{29\cdot30}+...+\frac{1}{73\cdot75}\)
\(C=\frac{4}{2\cdot4}+\frac{4}{4\cdot6}+\frac{4}{6\cdot8}+...3+\frac{4}{2008\cdot2010}\)
\(A=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(=7.\frac{1}{10.11}+7.\frac{1}{11.12}+7.\frac{1}{12.13}+...+7.\frac{1}{69.70}\)
\(=7.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)
\(=7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(=7.\left(\frac{1}{10}-\frac{1}{70}\right)=7.\frac{3}{35}=\frac{3}{5}\)
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\(A=7.\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(=\frac{6}{70}\)
\(=\frac{3}{35}\)
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\(B=\frac{1}{2}.\left(\frac{2}{25.27}+\frac{2}{27.29}+...+\frac{2}{73.75}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{75}\right)\)
\(=\frac{1}{2}.\frac{2}{75}\)
\(=\frac{1}{75}\)
=
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Tính bằng phương pháp hợp lí
a> C = \(\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+...+\frac{3}{73\cdot76}\) b> D = \(\frac{5}{10\cdot11}+\frac{5}{11\cdot12}+\frac{5}{12\cdot13}+...+\frac{5}{99\cdot100}\)
NHANH LÊN MÌNH CẦN GẤP
a) \(C=\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{73.76}\)
\(C=1.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{73}-\frac{1}{76}\right)\)
\(C=1.\left(\frac{1}{4}-\frac{1}{76}\right)\)
\(C=1.\frac{9}{38}\)
\(C=\frac{9}{38}\)
b) \(D=\frac{5}{10.11}+\frac{5}{11.12}+\frac{5}{12.13}+...+\frac{5}{99.100}\)
\(D=5.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{99}+\frac{1}{100}\right)\)
\(D=5.\left(\frac{1}{10}-\frac{1}{100}\right)\)
\(D=5.\frac{9}{100}\)
\(D=\frac{99}{20}\)
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\(\frac{1}{10\cdot11}\)+\(\frac{1}{11\cdot12}\)+\(\frac{1}{12\cdot13}\)+ ......................+\(\frac{1}{49\cdot50}\)
Ta có: \(\frac{1}{10.11}+\frac{1}{11.12}+....+\frac{1}{49.50}\)
\(\Rightarrow\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+....+\frac{1}{49}-\frac{1}{50}\)
\(=\frac{1}{10}-\frac{1}{50}\)
\(=\frac{2}{25}\)
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\(\frac{ }{\frac{4}{3\cdot7}+\frac{5}{7\cdot12}+\frac{1}{12\cdot13}+\frac{7}{13\cdot20}+\frac{3}{20\cdot23}}\)
làm cách thủ công đi
\(=\frac{20}{69}\)
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\(\frac{4}{3\cdot7}+\frac{5}{7\cdot12}+\frac{1}{12\cdot13}+\frac{7}{13\cdot20}+\frac{3}{20\cdot23}\)
\(\frac{4}{3.7}+\frac{5}{7.12}+\frac{1}{12.13}+\frac{7}{13.20}+\frac{3}{20.23}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+...+\frac{1}{20}-\frac{1}{23}=\frac{1}{3}-\frac{1}{23}=\frac{20}{69}\)
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\(\frac{4}{3\cdot7}+\frac{5}{7\cdot12}+\frac{1}{12\cdot13}+\frac{7}{13\cdot20}+\frac{8}{20\cdot28}\)
\(\frac{4}{3\cdot7}+\frac{5}{7.12}+..+\frac{8}{20\cdot28}\)
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+...+\frac{1}{20}-\frac{1}{28}\)
\(=\frac{1}{3}-\frac{1}{28}+0+...+0\)
\(=\frac{25}{84}\)
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\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+...+\frac{1}{20}-\frac{1}{28}\)
\(=\frac{1}{3}-\frac{1}{28}\)
\(=\frac{25}{84}\)
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Điền tiếp vào chỗ ... trong bài giải sau:Bài toán: Tìm x,y,znguyên dương biết frac{x}{7}+frac{y}{11}+frac{z}{13}0,left(946053right)Bài giải: Ta có: 0,left(946053right)frac{946053}{999999}frac{947}{1001}Rightarrowfrac{11cdot13x+13cdot7y+7cdot11z}{7cdot11cdot13}frac{947}{7cdot11cdot13}Rightarrow11cdot13x+13cdot7y+7cdot11z947Rightarrow143x+91y+77z947...Vậy hept{begin{cases}x3y4z2end{cases}}
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Điền tiếp vào chỗ ... trong bài giải sau:
Bài toán: Tìm \(x,y,z\)nguyên dương biết \(\frac{x}{7}+\frac{y}{11}+\frac{z}{13}=0,\left(946053\right)\)
Bài giải: Ta có: \(0,\left(946053\right)=\frac{946053}{999999}=\frac{947}{1001}\)
\(\Rightarrow\frac{11\cdot13x+13\cdot7y+7\cdot11z}{7\cdot11\cdot13}=\frac{947}{7\cdot11\cdot13}\)
\(\Rightarrow11\cdot13x+13\cdot7y+7\cdot11z=947\)
\(\Rightarrow143x+91y+77z=947\)
\(...\)
Vậy \(\hept{\begin{cases}x=3\\y=4\\z=2\end{cases}}\)
\(\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+\frac{4}{5\cdot7\cdot9}+\frac{4}{7\cdot9\cdot11}+\frac{4}{9\cdot11\cdot13}\)
giúp mk nha các bn
\(\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+\frac{4}{5\cdot7\cdot9}+\frac{4}{7\cdot9\cdot11}+\frac{4}{9\cdot11\cdot13}\)
\(=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{9.11}-\frac{1}{11.13}\)
\(=\frac{1}{1.3}-\frac{1}{11.13}\)
\(=\frac{1}{3}-\frac{1}{143}\)
\(=\frac{140}{429}\)
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