a) Đặt \(\left\{{}\begin{matrix}a=\frac{1}{2015}\\b=\frac{2011}{2013}\end{matrix}\right.\)
Ta có: \(D=\frac{4}{2015}\cdot\left(3+\frac{2011}{2013}\right)+\frac{1}{2015}\cdot\frac{2}{2013}-\frac{6033}{2013\cdot2015}\)
\(=4a\left(3+b\right)+a\left(1-b\right)-3ab\)
\(=12a+4ab+a-ab-3ab\)
\(=13a=13\cdot\frac{1}{2015}=\frac{13}{2015}\)
Vậy: \(D=\frac{13}{2015}\)
b) Ta có: \(\frac{1}{D}=1:\frac{13}{2015}\)
\(=1\cdot\frac{2015}{13}=\frac{2015}{13}\)
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