Question

1) A=7/10×11+7/11×12+7/12×13+…+7/69×70

2) B=1/25×27+1/27×29+1/29×31+…+1/73×75

3) C=4/2×4+4/4×6+4/6×8+…+4/2008×2010

Help me 😭 đang cần gấp

Answer 1

\(A=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)

\(A=7\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+....+\frac{1}{69.70}\right)\)

\(A=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+....+\frac{1}{69}-\frac{1}{70}\right)\)

\(A=7\left(\frac{1}{10}-\frac{1}{70}\right)\)

\(A=7\cdot\frac{3}{35}=\frac{21}{35}\)

Answer 2

\(A=\frac{7}{10\cdot11}+\frac{7}{11\cdot12}+\frac{7}{12\cdot13}+...+\frac{7}{69\cdot70}\)

\(A=7\left(\frac{1}{10\cdot11}+\frac{1}{11\cdot12}+\frac{1}{12\cdot13}+...+\frac{1}{69\cdot70}\right)\)

\(A=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)

\(A=7\left(\frac{1}{10}-\frac{1}{70}\right)=7\cdot\frac{3}{35}=\frac{3}{5}\)

\(B=\frac{1}{25\cdot27}+\frac{1}{27\cdot29}+\frac{1}{29\cdot31}+...+\frac{1}{73\cdot75}\)

\(B=\frac{1}{2}\left(\frac{2}{25\cdot27}+\frac{2}{27\cdot29}+\frac{2}{29\cdot31}+...+\frac{2}{73\cdot75}\right)\)

\(B=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\right)\)

\(B=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{75}\right)=\frac{1}{2}\cdot\frac{2}{75}=\frac{1}{75}\)

\(C=\frac{4}{2\cdot4}+\frac{4}{4\cdot6}+\frac{4}{6\cdot8}+...+\frac{4}{2008\cdot2010}\)

\(C=\frac{4}{2}\left(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{2008\cdot2010}\right)\)

\(C=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)

\(C=2\left(\frac{1}{2}-\frac{1}{2010}\right)=2\cdot\frac{502}{1005}=\frac{1004}{1005}\)

Answer 3

\(B=\frac{1}{25.27}+\frac{1}{27.29}+\frac{1}{29.31}+...+\frac{1}{73.75}\)

\(2B=\frac{2}{25.27}+\frac{2}{27.29}+\frac{1}{29.31}+...+\frac{1}{73.75}\)

\(2B=\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\)

\(2B=\frac{1}{25}-\frac{1}{75}=\frac{2}{75}\)

Answer 4

a) \(A=\frac{7}{10.11}+\frac{7}{11.12}+...+\frac{7}{69.70}=7\left(\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{69.70}\right)\)

\(=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)=7\left(\frac{1}{10}-\frac{1}{70}\right)=7.\frac{3}{35}=\frac{3}{5}\)

b) \(B=\frac{1}{25.27}+\frac{1}{27.29}+...+\frac{1}{73.75}=\frac{1}{2}\left(\frac{2}{25.27}+\frac{2}{27.29}+...+\frac{2}{73.75}\right)\)

\(=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\right)=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{75}\right)=\frac{1}{2}.\frac{2}{75}=\frac{1}{75}\)

c) \(C=\frac{4}{2.4}+\frac{4}{4.6}+...+\frac{4}{2008.2010}=2\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2008.2010}\right)\) 

\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2008}-\frac{1}{2010}\right)=2\left(\frac{1}{2}-\frac{1}{2010}\right)=2.\frac{502}{1005}=\frac{1004}{1005}\)

Answer 5

\(C=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{4.8}+...+\frac{4}{2008.2010}\)

\(2C=\frac{2}{2\cdot4}+\frac{2}{4.6}+...+\frac{2}{2008\cdot2010}\)

\(2C=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2008}-\frac{1}{2010}\)

\(C=2\left(\frac{1}{2}-\frac{1}{2010}\right)=2\cdot\frac{502}{1005}=\frac{1004}{1005}\)