\(\dfrac{3}{1.2.3.4.5}+\dfrac{3}{2.3.4.5.6}+\dfrac{3}{3.4.5.6.7}+...+\dfrac{3}{2011.2012.2013.2014.2015}\)
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Tính:
\(A=1.2.3.4.5+2.3.4.5.6+3.4.5.6.7+..........+46.47.48.49.50\)
A = 1.2.3.4.5 + 2.3.4.5.6 + 3.4.5.6.7 +...+ 46.47.48.49.50
6A = 1.2.3.4.5.(6-0) + 2.3.4.5.6.(7-1) + 3.4.5.6.7.(8-2) +...+ 46.47.48.49.50.(51-45)
6A = (1.2.3.4.5.6 + 2.3.4.5.6.7 + 3.4.5.6.7.8 +...+ 46.47.48.49.50.51) - (0.1.2.3.4.5 + 1.2.3.4.5.6 + 2.3.4.5.6.7 +...+ 45.46.47.48.49.50)
6A = 46.47.48.49.50.51 - 0.1.2.3.4.5
6A = 46.47.48.49.50.51
6A = 12966811200
A = 12966811200 : 6
A = 2161135200
k nha
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\(B=\dfrac{3}{3x5}+\dfrac{3}{5x7}+\dfrac{3}{7x9}+....+\dfrac{3}{48x50}\)Tính nhanh:
\(B=\dfrac{3}{3x5}+\dfrac{3}{5x7}+\dfrac{3}{7x9}+....+\dfrac{3}{48x50}\)
\(B=\dfrac{3}{3x5}+\dfrac{3}{5x7}+\dfrac{3}{7x9}+....+\dfrac{3}{48x50}\)
\(B=\dfrac{3}{3x5}+\dfrac{3}{5x7}+\dfrac{3}{7x9}+....+\dfrac{3}{48x50}\)
Giải:
\(B=\dfrac{3}{3\times5}+\dfrac{3}{5\times7}+\dfrac{3}{7\times9}+...+\dfrac{3}{48\times50}\)
\(B=\dfrac{3}{2}\times\left(\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+...+\dfrac{2}{48\times50}\right)\)
\(B=\dfrac{3}{2}\times\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{48}-\dfrac{1}{50}\right)\)
\(B=\dfrac{3}{2}\times\left(\dfrac{1}{3}-\dfrac{1}{50}\right)\)
\(B=\dfrac{3}{2}\times\dfrac{47}{150}\)
\(B=\dfrac{47}{100}\)
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3dfrac{3}{3}.dfrac{1}{3}-dfrac{3}{4}.dfrac{1}{3}left[dfrac{11}{3}right]-left(dfrac{-1}{2}right)^2-4dfrac{1}{2}left(dfrac{3}{2}-dfrac{5}{4}+dfrac{1}{3}right):left(dfrac{4}{3}+2dfrac{3}{2}-dfrac{3}{4}right)5dfrac{5}{27}+dfrac{7}{23}+0,5+dfrac{-5}{27}+dfrac{16}{23}2dfrac{5}{4}+left(-2018right)^0-left[dfrac{-1}{4}right]dfrac{19}{11}.dfrac{6}{5}+dfrac{6^2}{11}.dfrac{6}{5}-left(dfrac{1}{2}right)^0
Đọc tiếp
\(3\dfrac{3}{3}.\dfrac{1}{3}-\dfrac{3}{4}.\dfrac{1}{3}\)
\(\left[\dfrac{11}{3}\right]-\left(\dfrac{-1}{2}\right)^2-4\dfrac{1}{2}\)
\(\left(\dfrac{3}{2}-\dfrac{5}{4}+\dfrac{1}{3}\right):\left(\dfrac{4}{3}+2\dfrac{3}{2}-\dfrac{3}{4}\right)\)
\(5\dfrac{5}{27}+\dfrac{7}{23}+0,5+\dfrac{-5}{27}+\dfrac{16}{23}\)
\(2\dfrac{5}{4}+\left(-2018\right)^0-\left[\dfrac{-1}{4}\right]\)
\(\dfrac{19}{11}.\dfrac{6}{5}+\dfrac{6^2}{11}.\dfrac{6}{5}-\left(\dfrac{1}{2}\right)^0\)
\(\dfrac{0,75-0,6+\dfrac{3}{7}+\dfrac{3}{13}}{2,75-2,2+\dfrac{11}{7}+\dfrac{11}{3}}=\dfrac{\dfrac{3}{4}-\dfrac{3}{5}+\dfrac{3}{7}+\dfrac{3}{13}}{\dfrac{11}{4}-\dfrac{11}{5}+\dfrac{11}{7}+\dfrac{11}{13}}\)
22 tháng 7 2021 lúc 16:32
b) B=\(\dfrac{\dfrac{1}{22}+\dfrac{1}{13}-0,5}{\dfrac{3}{13}-\dfrac{3}{2}+\dfrac{3}{22}}.\dfrac{\dfrac{3}{4}-0,375+\dfrac{3}{16}-\dfrac{3}{32}}{1-\dfrac{1}{2}+\dfrac{1}{4}-0,875}+\dfrac{3}{4}\)
Ta có: \(B=\dfrac{\dfrac{1}{22}-\dfrac{1}{2}+\dfrac{1}{13}}{\dfrac{3}{22}-\dfrac{3}{2}+\dfrac{3}{13}}\cdot\dfrac{\dfrac{3}{4}-0.375+\dfrac{3}{16}-\dfrac{3}{32}}{1-\dfrac{1}{2}+\dfrac{1}{4}-0.875}+\dfrac{3}{4}\)
\(=\dfrac{1}{3}\cdot\dfrac{-15}{4}+\dfrac{3}{4}\)
\(=\dfrac{-5}{4}+\dfrac{3}{4}=\dfrac{-1}{2}\)
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1 tháng 3 2022 lúc 21:09
a) dfrac{3}{1.4} +dfrac{3}{4.7} + dfrac{3}{7.10} + ... + dfrac{3}{121.124}b) dfrac{3}{2.3} + dfrac{3}{3.4} + ... + dfrac{3}{100.101}c) dfrac{1}{1.5} + dfrac{1}{5.9} + dfrac{1}{9.13} + ... + dfrac{1}{401.405}d) dfrac{2}{1.3} + dfrac{2}{3.5} + dfrac{2}{5.7} + ... + dfrac{2}{99.101}
Đọc tiếp
a) \(\dfrac{3}{1.4}\) +\(\dfrac{3}{4.7}\) + \(\dfrac{3}{7.10}\) + ... + \(\dfrac{3}{121.124}\)
b) \(\dfrac{3}{2.3}\) + \(\dfrac{3}{3.4}\) + ... + \(\dfrac{3}{100.101}\)
c) \(\dfrac{1}{1.5}\) + \(\dfrac{1}{5.9}\) + \(\dfrac{1}{9.13}\) + ... + \(\dfrac{1}{401.405}\)
d) \(\dfrac{2}{1.3}\) + \(\dfrac{2}{3.5}\) + \(\dfrac{2}{5.7}\) + ... + \(\dfrac{2}{99.101}\)
a: \(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{121}-\dfrac{1}{124}=1-\dfrac{1}{124}=\dfrac{123}{124}\)
b: \(=3\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{100}-\dfrac{1}{101}\right)=3\cdot\dfrac{99}{202}=\dfrac{297}{202}\)
c: \(=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-...+\dfrac{1}{401}-\dfrac{1}{405}\right)=\dfrac{1}{4}\cdot\dfrac{404}{405}=\dfrac{101}{405}\)
d: \(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}=1-\dfrac{1}{101}=\dfrac{100}{101}\)
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Tính giá trị của biểu thức :\(\left(\dfrac{1}{200}+\dfrac{1}{300}+\dfrac{1}{400}+...+\dfrac{1}{1000}\right).1.2.3.4.5.\left(\dfrac{1}{120}-\dfrac{1}{180}-\dfrac{1}{360}\right)\)
\(\left(\dfrac{1}{200}+\dfrac{1}{300}+\dfrac{1}{400}+...+\dfrac{1}{1000}\right).1.2.3.4.5\).\(\left(\dfrac{1}{120}-\dfrac{1}{180}-\dfrac{1}{360}\right)\)
Xét \(\left(\dfrac{1}{120}-\dfrac{1}{180}-\dfrac{1}{360}\right)\) ta có :
= \(\dfrac{1}{120}-\left(\dfrac{1}{180}+\dfrac{1}{360}\right)\) =\(\dfrac{1}{120}-\dfrac{1}{120}=0\)
Trong 1 tích nếu có 1 thừa số 0 thì tích đó bằng 0
Biểu thức trên có 1 thừa số 0 nên biểu thức trên bằng 0
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18 tháng 3 2023 lúc 19:38
chứng minh rằng
a , \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+...+\dfrac{1}{512}-\dfrac{1}{1024}\) < \(\dfrac{1}{3}\)
b , \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\) < \(\dfrac{3}{16}\)
-\(\dfrac{4}{7}\)+\(\dfrac{15}{4}\)-(\(\dfrac{11}{4}\)+\(\dfrac{3}{7}+\dfrac{1}{2}-\dfrac{1}{3}\))
\(\dfrac{1}{5}\left(\dfrac{1}{2}-\dfrac{1}{3}\right):\left(\dfrac{-9}{10}\right)+\dfrac{-7}{3}\)
\(\dfrac{-20}{23}+\dfrac{2}{3}-\dfrac{3}{23}+\dfrac{2}{5}+\dfrac{7}{15}\)
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