Tính tổng : \(\dfrac{1}{5}+\dfrac{1}{20}+\dfrac{1}{44}+\dfrac{1}{77}+\dfrac{1}{119}+\dfrac{1}{170}+\dfrac{1}{230}+\dfrac{1}{299}\)
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5 tháng 2 2021 lúc 22:38
Tính tổng :a) B dfrac{1}{5} + dfrac{1}{20}+ dfrac{1}{44} +dfrac{1}{77} +dfrac{1}{119} + dfrac{1}{170} +dfrac{1}{230} +dfrac{1}{299} b) C left(1+dfrac{1}{1.3}right) left(1+dfrac{1}{2.4}right) left(1+dfrac{1}{3.5}right) .....left(1+dfrac{1}{2014.2016}right)
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Tính tổng :
a) B = \(\dfrac{1}{5}\) + \(\dfrac{1}{20}\)+ \(\dfrac{1}{44}\) +\(\dfrac{1}{77}\) +\(\dfrac{1}{119}\) + \(\dfrac{1}{170}\) +\(\dfrac{1}{230}\) +\(\dfrac{1}{299}\)
b) C = \(\left(1+\dfrac{1}{1.3}\right)\) \(\left(1+\dfrac{1}{2.4}\right)\) \(\left(1+\dfrac{1}{3.5}\right)\) .....\(\left(1+\dfrac{1}{2014.2016}\right)\)
Câu C giải rồi
\(B=\dfrac{1}{5}+\dfrac{1}{20}+\dfrac{1}{44}+\dfrac{1}{77}+\dfrac{1}{119}+\dfrac{1}{170}+\dfrac{1}{230}+\dfrac{1}{299}\)
\(=2\left(\dfrac{1}{10}+\dfrac{1}{40}+\dfrac{1}{88}+\dfrac{1}{154}+\dfrac{1}{238}+\dfrac{1}{340}+\dfrac{1}{460}+\dfrac{1}{598}\right)\)
\(=\dfrac{2}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+\dfrac{3}{14.17}+\dfrac{3}{17.20}+\dfrac{3}{20.23}+\dfrac{3}{23.26}\right)\)
\(=\dfrac{2}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{23}-\dfrac{1}{26}\right)\)
\(=\dfrac{2}{3}\left(\dfrac{1}{2}-\dfrac{1}{26}\right)=\dfrac{4}{13}\)
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Chứng minh A > B, biết:
A= \(\dfrac{2}{5.7}+\dfrac{5}{7.12}+\dfrac{7}{12.19}+\dfrac{9}{19.28}+\dfrac{11}{28.39}+\dfrac{1}{30.40}\)
B= \(\dfrac{1}{20}+\dfrac{1}{44}+\dfrac{1}{77}+\dfrac{1}{119}+\dfrac{1}{170}\)
Giúp mình với mình đang cần gấp!!!
Đây nha bạn:
=7−55.7+12−77.12+19−1212.19+28−1919.28+39−2828.39+40−3939.40
=15−17+17−112+112−119+119−128+128−139+139−140
=15−140=740
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Chứng minh A > B, biết A = \(\dfrac{2}{5.7}+\dfrac{5}{7.12}+\dfrac{7}{12.19}+\dfrac{9}{19.28}+\dfrac{11}{28.39}+\dfrac{1}{39.40}\)
B = \(\dfrac{1}{20}+\dfrac{1}{44}+\dfrac{1}{77}+\dfrac{1}{119}+\dfrac{1}{170}\)
ta tách 2/5x7 = 2/5-2/7 tách những cái kia tương tự góp vào rồi tính
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Tính giá trị các biểu thức sau theo cách hợp lý
a) A 3dfrac{1}{117}. 4 dfrac{1}{119} - 1 dfrac{116}{117} . 5 dfrac{118}{119} - dfrac{5}{119}
b) B 2dfrac{1}{315} . dfrac{1}{651} - dfrac{1}{105} .3dfrac{650}{651} - dfrac{4}{315.651} + dfrac{4}{105}
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Tính giá trị các biểu thức sau theo cách hợp lý
a) A= 3\(\dfrac{1}{117}\). 4 \(\dfrac{1}{119}\) - 1 \(\dfrac{116}{117}\) . 5 \(\dfrac{118}{119}\) - \(\dfrac{5}{119}\)
b) B= 2\(\dfrac{1}{315}\) . \(\dfrac{1}{651}\) - \(\dfrac{1}{105}\) .3\(\dfrac{650}{651}\) - \(\dfrac{4}{315.651}\) + \(\dfrac{4}{105}\)
Tính giá trị A= \(3\dfrac{1}{117}.\dfrac{1}{119}-\dfrac{4}{117}.5\dfrac{118}{119}-\dfrac{5}{117.119}+\dfrac{8}{39}\)
Ta có: \(A=3\dfrac{1}{117}\cdot\dfrac{1}{119}-\dfrac{4}{117}\cdot5\dfrac{118}{119}-\dfrac{5}{117\cdot119}+\dfrac{8}{39}\)
\(=\dfrac{352}{117}\cdot\dfrac{1}{119}-\dfrac{4}{117}\cdot\dfrac{713}{119}-\dfrac{5}{117\cdot119}+\dfrac{8}{39}\)
\(=\dfrac{352-2852-5}{117\cdot119}+\dfrac{8}{39}\)
\(=\dfrac{-835}{4641}+\dfrac{8}{39}\)
\(=\dfrac{3}{119}\)
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(\(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+\dfrac{1}{14\cdot19}+...+\dfrac{1}{44\cdot49}\)) .Tính tổng
=(54⋅9+59⋅14+514⋅19+...+544⋅49)⋅1−(3+5+7+...+49)89=(54⋅9+59⋅14+514⋅19+...+544⋅49)⋅1−(3+5+7+...+49)89=15⋅(14−19+19−114+114−119+...+144−149)⋅(1−(52⋅24)289)=15⋅(14−19+19−114+114−119+...+144−149)⋅(1−(52⋅24)289)=15⋅(14−149)⋅1−62489=15⋅(14−149)⋅1−62489=15⋅45196⋅−62389=15⋅45196⋅−62389=−928
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\(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+\dfrac{1}{14\cdot19}+...+\dfrac{1}{44\cdot49}\)
\(=\dfrac{1}{5}\left(\dfrac{9-4}{4\cdot9}+\dfrac{14-9}{9\cdot14}+\dfrac{19-14}{14\cdot19}+...+\dfrac{49-44}{44\cdot49}\right)\)
\(=\dfrac{1}{5}\cdot\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+....+\dfrac{1}{44}-\dfrac{1}{49}\right)\)
\(=\dfrac{1}{5}\cdot\left(\dfrac{1}{4}-\dfrac{1}{49}\right)\)
\(=\dfrac{1}{5}\cdot\dfrac{45}{196}\)
\(=\dfrac{9}{196}\)
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Tính giá trị biểu thức sau :
a) A = \(3\dfrac{1}{117}.4\dfrac{1}{119}-1\dfrac{116}{117}.5\dfrac{118}{119}-\dfrac{5}{119}\)
b) B = \(4\dfrac{1}{115}.3\dfrac{1}{225}-5\dfrac{114}{115}.1\dfrac{224}{225}-\dfrac{10}{225}\)
25 tháng 3 2022 lúc 21:31
Tính tổng: \(B=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\)
\(B=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\)
\(B=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}\)
\(B=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\)
\(B=1-\dfrac{1}{7}\)
\(B=\dfrac{6}{7}\)
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\(B=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\)
\(=\dfrac{6}{7}\)
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Xem thêm câu trả lời
a) Tính :
\(1-\dfrac{1}{2};\dfrac{1}{2}-\dfrac{1}{3};\dfrac{1}{3}-\dfrac{1}{4};\dfrac{1}{4}-\dfrac{1}{5};\dfrac{1}{5}-\dfrac{1}{6}\)
b) Sử dụng kết quả của câu a) để tính nhanh tổng sau :
\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}\)
a) \(1-\dfrac{1}{2}=\dfrac{1}{2}\)
\(\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{3-2}{6}=\dfrac{1}{6}\)
\(\dfrac{1}{3}-\dfrac{1}{4}=\dfrac{4-3}{12}=\dfrac{1}{12}\)
\(\dfrac{1}{4}-\dfrac{1}{5}=\dfrac{5-4}{20}=\dfrac{1}{20}\)
\(\dfrac{1}{5}-\dfrac{1}{6}=\dfrac{6-5}{30}=\dfrac{1}{30}\)
b) \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}\)
\(=\left(1-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+\left(\dfrac{1}{3}-\dfrac{1}{4}\right)+\left(\dfrac{1}{4}-\dfrac{1}{5}\right)+\left(\dfrac{1}{5}+\dfrac{1}{6}\right)\)
\(=1+\left(-\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(-\dfrac{1}{3}+\dfrac{1}{3}\right)+\left(-\dfrac{1}{4}+\dfrac{1}{4}\right)+\left(-\dfrac{1}{5}+\dfrac{1}{5}\right)+-\dfrac{1}{6}\)\(=1+-\dfrac{1}{6}\)
\(=\dfrac{5}{6}\)
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