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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: \(\frac{227}{261}\)
\(A=\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)...\left(1-\dfrac{1}{2016^2}\right)\)
\(=\left[\left(1\right)^2-\left(\dfrac{1}{2}\right)^2\right]\left[\left(1\right)^2-\left(\dfrac{1}{3}\right)^2\right]...\left[\left(1\right)^2-\left(\dfrac{1}{2016}\right)^2\right]\)
\(=\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)...\left(1-\dfrac{1}{2016}\right)\left(1+\dfrac{1}{2016}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{3}{2}\cdot\dfrac{2}{3}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{2015}{2016}\cdot\dfrac{2017}{2016}\)
\(=\dfrac{1}{2}\cdot\dfrac{2017}{2016}=\dfrac{2017}{4032}\)
\(C=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+..+\frac{1}{2016}.\left(1+2+3+...+2016\right)\)
\(C=1+\frac{1}{2}.\left(1+2\right).2:2+\frac{1}{3}.\left(1+3\right).3:2+\frac{1}{4}.\left(1+4\right).4:2+...+\frac{1}{2016}.\left(1+2016\right).2016:2\)
\(C=1+3:2+4:2+5:2+...+2017:2\)
\(C=2.\frac{1}{2}+3.\frac{1}{2}+4.\frac{1}{2}+5.\frac{1}{2}+...+2017.\frac{1}{2}\)
\(C=\frac{1}{2}.\left(2+3+4+5+...+2017\right)\)
\(C=\frac{1}{2}.\left(2+2017\right).2016:2\)
\(C=\frac{1}{2}.2019.2016.\frac{1}{2}\)
\(C=2019.504=1017576\)