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\(A=\dfrac{1}{2}-\left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{2}\right)^3-...-\left(\dfrac{1}{2}\right)^{10}\\ \\2A=1-\dfrac{1}{2}-\left(\dfrac{1}{2}\right)^2-...-\left(\dfrac{1}{2}\right)^9\\ 2A-A=\left[1-\dfrac{1}{2}-\left(\dfrac{1}{2}\right)^2-...-\left(\dfrac{1}{2}\right)^9\right]-\left[\dfrac{1}{2}-\left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{2}\right)^3-...-\left(\dfrac{1}{2}\right)^{10}\right]\\ A=1-\dfrac{1}{4}+\left(\dfrac{1}{2}\right)^{10}\)
\(D=-\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)\cdot...\cdot\left(1-\frac{1}{100^2}\right).\)
\(D=-\frac{2^2-1}{2^2}\cdot\frac{3^2-1}{3^2}\cdot\frac{4^2-1}{4^2}\cdot...\cdot\frac{100^2-1}{100^2}.\)
\(D=-\frac{1\cdot3}{2^2}\cdot\frac{2\cdot4}{3^2}\cdot\frac{3\cdot5}{4^2}\cdot\frac{4\cdot6}{5^2}\cdot...\cdot\frac{98\cdot100}{99^2}\cdot\frac{99\cdot101}{100^2}=-\frac{1}{2}\cdot\frac{101}{100}=-\frac{101}{200}\)
\(A=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)...\left(\frac{1}{2002}-1\right)\left(\frac{1}{2003}-1\right)\)
\(=\left(-\frac{1}{2}\right)\left(-\frac{2}{3}\right)...\left(-\frac{2001}{2002}\right)\left(-\frac{2002}{2003}\right)\)
\(=\frac{-1.\left(-2\right).....\left(-2001\right)\left(-2002\right)}{2.3....2002.2003}\)
\(=\frac{1}{2003}\)
a) \(A=\left(1:\frac{1}{4}\right).4+25\left(1:\frac{16}{9}:\frac{125}{64}\right):\left(-\frac{27}{8}\right)\)
\(=4.4+25.\frac{36}{125}:\frac{-27}{8}\)
\(=16-\frac{32}{15}=\frac{240}{15}-\frac{32}{15}=\frac{208}{15}\)
\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...............+\frac{1}{500}\left(1+2+3+.........+500\right)\)
\(=1+\frac{1}{2}\frac{3.2}{2}+\frac{1}{3}\frac{4.3}{2}+.............+\frac{1}{500}\frac{501.500}{2}\)
\(=\frac{1}{2}\left(2+3+............+501\right)\)
\(=\frac{1}{2}.251000\)
\(=125500\)