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\(=\dfrac{1}{2}\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{13\cdot15}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{4}{15}=\dfrac{2}{15}\)
a)\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)
\(=\frac{1}{2}.\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+\frac{11-9}{9.11}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{11}\right)\)
\(\frac{10}{22}\)
\(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\)
\(=2\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\right)\)
\(=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\)
\(=\dfrac{1}{3}-\dfrac{1}{13}\)
\(=\dfrac{10}{39}\)
nhưng khi tính trong máy tính được kết quả là \(\dfrac{5}{39}\) mà bạn
`a)1/2 . [-3]/4 . [-5]/8 . [-8]/9=[1. (-3).(-5).(-8)]/[2.4.8.3.3]=[-5]/[2.4.3]=[-5]/24`
`b)(2/[1.3]+2/[3.5]+2/[5.7]).([10.13]/3-[2^2]/3-[5^3]/3)`
`=(1-1/3+1/3-1/5+1/5-1/7).[10.13-2^2-5^3]/3`
`=(1-1/7).[130-4-125]/3`
`=6/7 . 1/3 = 2/7`
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`8/9+1/9 . 2/9+1/9 . 7/9`
`=8/9+1/9.(2/9+7/9)`
`=8/9+1/9 . 9/9`
`=8/9+1/9=9/9=1`
a) \(\dfrac{1}{2}\cdot\dfrac{-3}{4}\cdot\dfrac{-5}{8}\cdot\dfrac{-8}{9}\)
\(=\dfrac{1\cdot\left(-3\right)\cdot\left(-5\right)\cdot\left(-8\right)}{2\cdot4\cdot8\cdot9}\)
\(=-\dfrac{5}{24}\)
b) \(\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}\right)\cdot\left(\dfrac{10\cdot13}{3}-\dfrac{2^2}{3}-\dfrac{5^3}{3}\right)\)
\(=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}\right)\cdot\left(\dfrac{130}{3}-\dfrac{4}{3}-\dfrac{125}{3}\right)\)
\(=\left(1-\dfrac{1}{7}\right)\cdot\dfrac{1}{3}\)
\(=\dfrac{6}{7}\cdot\dfrac{1}{3}\)
\(=\dfrac{2}{7}\)
\(\dfrac{8}{9}+\dfrac{1}{9}\cdot\dfrac{2}{9}+\dfrac{1}{9}\cdot\dfrac{7}{9}\)
\(=\dfrac{8}{9}+\dfrac{2}{81}+\dfrac{7}{81}\)
\(=\dfrac{72}{81}+\dfrac{2}{81}+\dfrac{7}{81}\)
\(=1\)
\(M=\dfrac{6}{2.5}+\dfrac{6}{5.8}+\dfrac{6}{8.11}+...+\dfrac{6}{47.50}\)
\(\Rightarrow\dfrac{M}{2}=\dfrac{6:2}{2.5}+...+\dfrac{6:2}{47.50}\)
\(=\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{47.50}\)
\(=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{47}-\dfrac{1}{50}\)
\(=\dfrac{1}{2}-\dfrac{1}{50}\)
\(=\dfrac{12}{25}\)
\(\Rightarrow M=\dfrac{12}{25}.2=\dfrac{24}{25}\)
\(K=\dfrac{1}{9.11}+\dfrac{1}{11.13}+\dfrac{1}{13.15}+...+\dfrac{1}{43.45}\)
\(\Rightarrow2K=\dfrac{2}{9.11}+\dfrac{2}{11.13}+\dfrac{2}{13.15}+...+\dfrac{2}{43.45}\)
\(=\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{43}-\dfrac{1}{45}\)
\(=\dfrac{1}{9}-\dfrac{1}{45}\)
\(=\dfrac{4}{45}\)
\(\Rightarrow K=\dfrac{4}{45}:2=\dfrac{2}{45}\)
\(M=\dfrac{6}{2.5}+\dfrac{6}{5.8}+\dfrac{6}{8.11}+...+\dfrac{6}{47.50}\)
\(M=\dfrac{6}{3}.\left(\dfrac{6}{2}-\dfrac{6}{5}+\dfrac{6}{5}-\dfrac{6}{8}+\dfrac{6}{8}-\dfrac{6}{11}+...+\dfrac{6}{47}-\dfrac{6}{50}\right)\)
\(M=\dfrac{6}{3}.\left(\dfrac{6}{2}-\dfrac{6}{50}\right)\)
\(M=\dfrac{6}{3}.\left(\dfrac{150}{50}-\dfrac{6}{50}\right)\)
\(M=\dfrac{6}{3}.\dfrac{144}{50}\)
\(M=\dfrac{144}{25}\)
\(K=\dfrac{1}{9.11}+\dfrac{1}{11.13}+\dfrac{1}{13.15}+...+\dfrac{1}{43.45}\)
\(K=\dfrac{1}{2}.\left(\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{43}-\dfrac{1}{45}\right)\)
\(K=\dfrac{1}{2}.\left(\dfrac{1}{9}-\dfrac{1}{45}\right)\)
\(K=\dfrac{1}{2}.\left(\dfrac{5}{45}-\dfrac{1}{45}\right)\)
\(K=\dfrac{1}{2}.\dfrac{4}{45}\)
\(K=\dfrac{2}{45}\)
a, A= \(5\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)\)
\(A=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(A=5\left(1-\dfrac{1}{100}\right)\)
\(A=5.\dfrac{99}{100}=\dfrac{99}{20}.\)
b, \(C=1.2.3+2.3.4+...+8.9.10\)
\(4C=1.2.3.4+2.3.4.\left(5-1\right)+...+8.9.10.\left(11-7\right)\)\(4C=1.2.3.4+2.3.4.5-1.2.3.4+...+8.9.10.11-7.8.9.10\)\(4C=8.9.10.11\)
\(C=\dfrac{8.9.10.11}{4}=1980.\)
c, https://hoc24.vn/hoi-dap/question/384591.html
Câu này bạn vào đây mình đã giải câu tương tự nhé.
\(1)A=\dfrac{5}{1.2}+\dfrac{5}{2.3}+...+\dfrac{5}{99.100}\)
\(\Leftrightarrow A=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(\Leftrightarrow A=5\left(1-\dfrac{1}{100}\right)\)
\(\Leftrightarrow A=5\cdot\dfrac{99}{100}\)
\(\Leftrightarrow A=\dfrac{99}{20}\)
Xét C = \(\left(\dfrac{4}{1.3}+\dfrac{4}{3.5}+...+\dfrac{4}{97.99}\right)-\left(\dfrac{5}{11.12}+\dfrac{5}{12.13}+...+\dfrac{5}{98.99}\right)\)
Đặt A = \(\dfrac{4}{1.3}+\dfrac{4}{3.5}+...+\dfrac{4}{97.99}\)
B = \(\dfrac{5}{11.12}+\dfrac{5}{12.13}+...+\dfrac{5}{98.99}\)
=> C = A - B
Ta có : A = \(\dfrac{4}{1.3}+\dfrac{4}{3.5}+...+\dfrac{4}{97.99}\)
= 2 \(\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{97.99}\right)\)
= \(2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
= \(2\left(1-\dfrac{1}{99}\right)=\dfrac{2.98}{99}=\dfrac{196}{99}\)
Ta có B = \(\dfrac{5}{11.12}+\dfrac{5}{12.13}+...+\dfrac{5}{98.99}\)
= \(5\left(\dfrac{1}{11.12}+\dfrac{1}{12.13}+...+\dfrac{1}{98.99}\right)\)
= \(5\left(\dfrac{1}{11}-\dfrac{1}{99}\right)=\dfrac{8.5}{99}=\dfrac{40}{99}\)
=> C = A - B = \(\dfrac{196-40}{99}=\dfrac{156}{99}=\dfrac{52}{33}\)
\(C=\dfrac{4}{1.3}+\dfrac{4}{3.5}+.....+\dfrac{4}{97.99}-\dfrac{5}{11.12}-\dfrac{5}{12.13}-.....-\dfrac{5}{98.99}\)
\(C=\left(\dfrac{4}{1.3}+\dfrac{4}{3.5}+.....+\dfrac{4}{97.99}\right)-\left(\dfrac{5}{11.12}+\dfrac{5}{12.13}+.....+\dfrac{5}{98.99}\right)\)\(C=2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+.....+\dfrac{1}{97}-\dfrac{1}{99}\right)-5\left(\dfrac{1}{11}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{13}+.....+\dfrac{1}{98}-\dfrac{1}{99}\right)\)\(C=2\left(1-\dfrac{1}{99}\right)-5\left(\dfrac{1}{11}-\dfrac{1}{99}\right)\)
\(C=2.\dfrac{98}{99}-5.\dfrac{8}{99}\)
\(C=\dfrac{196}{99}-\dfrac{40}{99}=\dfrac{52}{33}\)
\(\dfrac{1}{1.5}+\dfrac{1}{5.9}+...+\dfrac{1}{37.41}\)
\(=\dfrac{1}{4}\left(\dfrac{4}{1.5}+\dfrac{4}{5.9}+...+\dfrac{4}{37.41}\right)\)
\(=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{37}-\dfrac{1}{41}\right)\)
\(=\dfrac{1}{4}.\left(1-\dfrac{1}{41}\right)\)
\(=\dfrac{1}{4}.\dfrac{40}{41}\)
\(=\dfrac{10}{41}\)