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Đặt : \(A=10.11+11.12+...+98.99+99.100\)
\(\Rightarrow3A=10.11.3+11.12.3+...+98.99.3+99.100.3\)
\(\Rightarrow3A=10.11.\left(12-9\right)+11.12.\left(13-10\right)+...+\)\(98.99.\left(100-97\right)+99.100.\left(101-98\right)\)
Số số hạng là(99,100-10,11):0,01+1=8900
=>Tổng=8900:2x(10,11+99,100)=485984,5
\(A=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(\Rightarrow A=7.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+....+\frac{1}{69.70}\right)\)
\(\Rightarrow A=7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(\Rightarrow A=7\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(\Rightarrow A=7.\frac{3}{35}\)
\(\Rightarrow A=\frac{3}{5}\)
A= 5/10.11+5/11.12+5/12.13+5/13.14+5/14.15
A= (1/10.11+1/11.12+1/12.13+1/13.14+1/14.15) :5
A= [(1/10-1/11)+(1/11-1/12)+(1/12-1/13)+(1/13-1/14)+(1/14-1/15)] :5
A= (1/10-1/15):5
A= 1/30:5
A= 1/150
\(C=\frac{7}{10.11}+\frac{7}{11.12}+...+\frac{7}{69.70}\)
\(\Leftrightarrow C=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(\Leftrightarrow C=7\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(\Leftrightarrow C=7\cdot\frac{3}{35}\)
\(\Leftrightarrow C=\frac{3}{5}\)
C=7(1/10.11+1/11.12+...+1/69.70)
=7(1/10-1/11+1/11-1/12+...+1/69-1/70)
=7(1/10-1/70)
=7.3/35
=3/5
Tính :
a) \(M=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{100}{101}\)
b) \(A=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(=7.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)
\(=7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(=7.\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(=7.\frac{3}{35}\)
\(=\frac{3}{5}\)
c) \(B=\frac{1}{25.27}+\frac{1}{27.29}+\frac{1}{29.31}+...+\frac{1}{73.75}\)
\(=\frac{1}{2}.\left(\frac{2}{25.27}+\frac{2}{27.29}+\frac{2}{29.31}+...+\frac{2}{73.75}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{75}\right)\)
\(=\frac{1}{2}.\frac{2}{75}\)
\(=\frac{1}{75}\)
\(\frac{A}{2}=\frac{1}{10\cdot11}+\frac{1}{11\cdot12}+...+\frac{1}{99\cdot100}\)
\(\frac{A}{2}=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{99}-\frac{1}{100}\)
\(\frac{A}{2}=\frac{1}{10}-\frac{1}{100}\)
\(\frac{A}{2}=\frac{9}{100}\)
\(A=\frac{9}{50}\)
\(A=\frac{2}{10\cdot11}+\frac{2}{11\cdot12}+\frac{2}{12\cdot13}+...+\frac{2}{99\cdot100}\)
\(A=2\left(\frac{1}{10\cdot11}+\frac{1}{11\cdot12}+\frac{1}{12\cdot13}+...+\frac{1}{99\cdot100}\right)\)
\(A=2\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=2\left(\frac{1}{10}-\frac{1}{100}\right)\)
\(A=2\cdot\frac{9}{100}\)
\(A=\frac{9}{50}\)