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NH
3
HM
0
25 tháng 4 2018
A = \(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
A=\(\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{50}{100}-\dfrac{1}{100}=\dfrac{49}{100}\)
25 tháng 4 2018
B = \(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{49.51}\)
B = \(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{49}-\dfrac{1}{51}\)
B = \(\dfrac{1}{2}-\dfrac{1}{51}=\dfrac{51}{102}-\dfrac{2}{102}=\dfrac{49}{102}\)
IM
20 tháng 8 2016
b)
\(\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
\(=\frac{32}{99}\)
c)
\(\frac{7}{3.4}+\frac{7}{4.5}+.....+\frac{7}{60.61}\)
\(=7\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....+\frac{1}{60}-\frac{1}{61}\right)\)
\(=7\left(\frac{1}{3}-\frac{1}{61}\right)\)
\(=\frac{406}{183}\)
d)
\(\frac{6}{2.4}+\frac{6}{4.6}+....+\frac{1}{72.74}\)
\(=3\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+.....+\frac{1}{72}-\frac{1}{74}\right)\)
\(=3\left(\frac{1}{2}-\frac{1}{74}\right)\)
=57/37
SM
1
7 tháng 8 2019
b) \(\frac{-5\cdot7^5+7^4}{7^6\cdot10-2\cdot7^5}\)
\(=\frac{-35\cdot7^4+7^4}{7^5\cdot70-2\cdot7^5}\)
\(=\frac{7^4\left(-35+1\right)}{7^5\left(70-2\right)}\)
\(=\frac{7^4\cdot\left(-34\right)}{7^5\cdot68}\)
\(=\frac{-1}{14}\)
Chắc sai =))
TT
1
VA
25 tháng 7 2016
Ta có :
\(\frac{7^5.6^8.\left(-3\right)^7}{9^2.7^4.6^9}\)\(=\frac{7^4.7.6^8.\left(-1.3\right)^7}{\left(3^2\right)^2.7^4.6^8.6}=\frac{7^4.7.6^8\left(-1\right)^7.3^4.3^3}{3^4.7^4.6^8.6}=\frac{7.\left(-1\right).27}{6}=-\frac{7.27}{6}=-\frac{189}{6}=-\frac{63}{2}\)
Ủng hộ mk nha !!! ^_^
6 tháng 6 2015
A = \(\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
=\(7\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)
=\(7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\right)\)
=\(7\left(\frac{1}{10}-\frac{1}{70}\right)\)
=\(7.\frac{3}{35}\)
=\(\frac{3}{5}\)
B=\(\frac{1}{25.27}+\frac{1}{27.29}+\frac{1}{29.31}+...+\frac{1}{73.75}\)
=\(\frac{1}{2}\left(\frac{2}{25.27}+\frac{2}{27.29}+\frac{2}{29.31}+...+\frac{2}{73.75}\right)\)
=\(\frac{1}{2}\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\right)\)
=\(\frac{1}{2}\left(\frac{1}{25}-\frac{1}{75}\right)\)
=\(\frac{1}{2}.\frac{2}{75}\)
=\(\frac{1}{75}\)
T
0
A =\(\dfrac{7}{3.4}\) + \(\dfrac{7}{4.6}\) + \(\dfrac{7}{5.8}\) + \(\dfrac{7}{6.10}\)+...+\(\dfrac{7}{60.118}\)
A = \(\dfrac{2.7}{2.3.4}\) + \(\dfrac{2.7}{2.4.6}\)+\(\dfrac{2.7}{2.5.8}\) + \(\dfrac{2.7}{2.6.10}\)+...+\(\dfrac{2.7}{2.60.118}\)
A = 7.(\(\dfrac{2}{6.4}\)+\(\dfrac{2}{8.6}\)+\(\dfrac{2}{10.8}\)+\(\dfrac{2}{12.10}\)+...+\(\dfrac{2}{120.118}\))
A = 7.(\(\dfrac{2}{4.6}\)+\(\dfrac{2}{6.8}\)+\(\dfrac{2}{8.10}\)+\(\dfrac{2}{10.12}\)+...+\(\dfrac{2}{118.120}\))
A = 7.(\(\dfrac{1}{4}-\dfrac{1}{6}\)+ \(\dfrac{1}{6}-\dfrac{1}{8}\) +\(\dfrac{1}{8}\) - \(\dfrac{1}{10}\) + \(\dfrac{1}{10}\) - \(\dfrac{1}{12}\) +...+ \(\dfrac{1}{118}\) - \(\dfrac{1}{120}\))
A = 7.( \(\dfrac{1}{4}\) - \(\dfrac{1}{120}\))
A = 7.\(\dfrac{29}{120}\)
A = \(\dfrac{203}{120}\)