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9 tháng 2 2019

\(\Leftrightarrow5^3\times\left(-2\right)^3\times\left(-61\right)\times1\)\(1\)

\(=[5\times\left(-2\right)]^3\times\left(-61\right)\)

\(=\left(-10\right)^3\times\left(-61\right)\)

\(=\left(-1000\right)\times\left(-61\right)\)

\(=61000\)

25 tháng 11 2017

Cac ban lam nhanh gium nhe

10 tháng 3 2017

\(D= \dfrac{1}{1.3} + \dfrac{1}{3.5}+...+\dfrac{1}{\left(2n-1\right).\left(2n+1\right)}\),

\(2.D = \dfrac{2}{1.3}+ \dfrac{2}{3.5}+...+\dfrac{2}{\left(2n-1\right).\left(2n+1\right)}\)

\(2.D = 1 - \dfrac{1}{3} + \dfrac{1}{3}- \dfrac{1}{5} +\dfrac{1}{5}- \dfrac{1}{7} + ... + \dfrac{1}{\left(2n-1\right)}-\dfrac{1}{\left(2n+1\right)}\)

\(2.D = 1 - \dfrac{1}{\left(2n+1\right)}\)

\(2.D= \dfrac{2n}{\left(2n+1\right)} \)

Vậy \(D = \dfrac{n}{\left(2n+1\right)}\)

10 tháng 3 2017

\(E=\dfrac{1}{1.3.5}+\dfrac{1}{3.5.7}+\dfrac{1}{5.7.9}+...+\dfrac{1}{\left(2n-1\right).\left(2n+1\right).\left(2n+3\right)}\)

\(\Rightarrow4E=4.\dfrac{1}{1.3.5}+\dfrac{1}{3.5.7}+\dfrac{1}{5.7.9}+...+\dfrac{1}{\left(2n-1\right).\left(2n+1\right).\left(2n+3\right)}\)

\(=\dfrac{4}{1.3.5}+\dfrac{4}{3.5.7}+...+\dfrac{4}{\left(2n-1\right).\left(2n+1\right).\left(2n+3\right)}\)

\(=\dfrac{1}{1.3}-\dfrac{1}{3.5}+\dfrac{1}{3.5}-\dfrac{1}{5.7}-...+\dfrac{1}{\left(2n-1\right).\left(2n+1\right)}-\dfrac{1}{\left(2n+1\right).\left(2n+3\right)}\)

\(=\dfrac{1}{1.3}-\dfrac{1}{\left(2n+1\right).\left(2n+3\right)}\)

\(\Rightarrow E=\dfrac{\dfrac{1}{1.3}-\dfrac{1}{\left(2n+1\right).\left(2n+3\right)}}{4}\)

\(=\dfrac{1}{12}-\dfrac{1}{\left(2n+1\right).\left(2n+3\right).4}\)