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\(\frac{\left(-\frac{1}{3}\right)^2-\left(\frac{3}{4}\right)^3.\left(-2\right)^2}{2.\left(-1\right)^5+\left(\frac{3}{4}\right)^2-\frac{3}{8}}=\frac{\frac{1}{9}-\frac{27}{64}.4}{-2+\frac{9}{16}-\frac{3}{8}}=\frac{\frac{1}{9}-\frac{27}{16}}{-2+\frac{3}{16}}\)
\(=\frac{\frac{16}{144}-\frac{243}{144}}{-\frac{32}{16}+\frac{3}{16}}=\frac{\frac{-227}{144}}{\frac{-29}{16}}=\frac{-227}{144}.\frac{-16}{29}\)
\(=\frac{227.16}{144.29}=\frac{227.1}{9.29}=\frac{227}{261}\)
Đáp số: \(\frac{227}{261}\)
\(A=\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)...\left(1-\dfrac{1}{2016^2}\right)\)
\(=\left[\left(1\right)^2-\left(\dfrac{1}{2}\right)^2\right]\left[\left(1\right)^2-\left(\dfrac{1}{3}\right)^2\right]...\left[\left(1\right)^2-\left(\dfrac{1}{2016}\right)^2\right]\)
\(=\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)...\left(1-\dfrac{1}{2016}\right)\left(1+\dfrac{1}{2016}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{3}{2}\cdot\dfrac{2}{3}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{2015}{2016}\cdot\dfrac{2017}{2016}\)
\(=\dfrac{1}{2}\cdot\dfrac{2017}{2016}=\dfrac{2017}{4032}\)