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\(A=\dfrac{10}{7.12}+\dfrac{10}{12.17}+\dfrac{10}{17.22}+...+\dfrac{10}{502.507}\) (sửa 502+507 thành 503.507)
\(\Rightarrow A=10\left(\dfrac{1}{7.12}+\dfrac{1}{12.17}+\dfrac{1}{17.22}+...+\dfrac{1}{502.507}\right)\)
\(\Rightarrow A=10.\dfrac{1}{5}\left(\dfrac{1}{7}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{22}+...+\dfrac{1}{502}-\dfrac{1}{507}\right)\)
\(\Rightarrow A=2.\left(\dfrac{1}{7}-\dfrac{1}{507}\right)=2.\left(\dfrac{500}{3549}\right)=\dfrac{1000}{3549}\)
\(B=\dfrac{4}{8.13}+\dfrac{4}{13.18}+\dfrac{4}{18.23}+...+\dfrac{4}{253.258}\)
\(\Rightarrow B=4\left(\dfrac{1}{8.13}+\dfrac{1}{13.18}+\dfrac{1}{18.23}+...+\dfrac{1}{253.258}\right)\)
\(\Rightarrow B=4.\dfrac{1}{5}\left(\dfrac{1}{8}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{18}+\dfrac{1}{18}-\dfrac{1}{23}+...+\dfrac{1}{253}-\dfrac{1}{258}\right)\)
\(\Rightarrow B=\dfrac{4}{5}\left(\dfrac{1}{8}-\dfrac{1}{258}\right)=\dfrac{4}{5}\left(\dfrac{129}{1032}-\dfrac{8}{1032}\right)=\dfrac{4}{5}.\dfrac{121}{1032}=\dfrac{121}{1290}\)
3.đặt biểu thức trên là A
ta có:
2A=2.(\(\dfrac{1}{3.5}\)+\(\dfrac{1}{5.7}\)+\(\dfrac{1}{7.9}\)+...+\(\dfrac{1}{2015.2017}\))
=>2A=\(\dfrac{2}{3.5}\)+\(\dfrac{2}{5.7}\)+\(\dfrac{2}{7.9}\)+....+\(\dfrac{2}{2015.2017}\)
=>2A=\(\dfrac{1}{3}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{9}\)+....\(\dfrac{1}{2015}\)-\(\dfrac{1}{2017}\)
=>2A=\(\dfrac{1}{3}\)-\(\dfrac{1}{2017}\)=\(\dfrac{2014}{6051}\)
=>A=\(\dfrac{2014}{6051}\):2=\(\dfrac{1007}{6051}\)
a) \(\dfrac{1}{2}.\dfrac{1}{-3}+\dfrac{1}{-3}.\dfrac{1}{4}+\dfrac{1}{4}.\dfrac{1}{-5}+\dfrac{1}{-5}.\dfrac{1}{6}\)
\(=\dfrac{1}{-3}\left(\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{1}{-5}\left(\dfrac{1}{4}+\dfrac{1}{6}\right)\)
\(=\dfrac{1}{-3}.\dfrac{3}{4}+\dfrac{1}{-5}.\dfrac{5}{12}\)
\(=\left(-\dfrac{1}{4}\right)+\left(-\dfrac{1}{12}\right)\)
\(=-\dfrac{1}{3}\)
b) \(A=\dfrac{81^4.3^{10}.27^5.3^{12}}{3^{18}.9^3.243^2}\)
\(=\dfrac{9^8.9^8.9^{13}.9^{10}}{9^{16}.9^3.9^3}\)
\(=\dfrac{9^{39}}{9^{22}}\)
\(=9^{17}\)
\(A=\dfrac{81^4\cdot3^{10}\cdot27^5\cdot3^{12}}{3^{18}\cdot9^3\cdot243^2}=\dfrac{3^{16}\cdot3^{10}\cdot3^{15}\cdot3^{12}}{3^{18}\cdot3^6\cdot3^{10}}=\dfrac{3^{53}}{3^{34}}=3^{19}\)
Vậy A = 319
Ngân Hà làm đúng phần a) nhưng làm sai phần b) nên mk chỉ làm phần b) thôi
=(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
\(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+\dfrac{1}{14\cdot19}+...+\dfrac{1}{44\cdot49}\)
\(=\dfrac{1}{5}\left(\dfrac{9-4}{4\cdot9}+\dfrac{14-9}{9\cdot14}+\dfrac{19-14}{14\cdot19}+...+\dfrac{49-44}{44\cdot49}\right)\)
\(=\dfrac{1}{5}\cdot\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+....+\dfrac{1}{44}-\dfrac{1}{49}\right)\)
\(=\dfrac{1}{5}\cdot\left(\dfrac{1}{4}-\dfrac{1}{49}\right)\)
\(=\dfrac{1}{5}\cdot\dfrac{45}{196}\)
\(=\dfrac{9}{196}\)
