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Lời giải:
Ta có: \(D=(2!)^2\left(\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+....+\frac{1}{2015^2}\right)\)
Xét số hạng tổng quát dạng \(\frac{1}{(2n+1)^2}\) với \(n\in\mathbb{N}\ge 1\)
Ta có: \((2n+1)^2-2n(2n+2)=1>0\)
\(\Rightarrow (2n+1)^2> 2n(2n+2)\Rightarrow \frac{1}{(2n+1)^2}< \frac{1}{2n(2n+2)}\)
Do đó: \(\left\{\begin{matrix} \frac{1}{3^2}< \frac{1}{2.4}\\ \frac{1}{5^2}< \frac{1}{4.6}\\ .....\\ \frac{1}{2015^2}< \frac{1}{2014.2016}\end{matrix}\right.\)
\(\Rightarrow \frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+...+\frac{1}{2015^2}< 1+\frac{1}{2.4}+\frac{1}{4.6}+....+\frac{1}{2014.1016}\)
\(\Leftrightarrow \frac{D}{(2!)^2}< 1+\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+....+\frac{1}{2014.2016}\)
\(\Leftrightarrow D< 4\left(1+\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+....+\frac{1}{2014.2016}\right)\)
\(\Leftrightarrow D< 4+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1007.1008}\)
\(\Leftrightarrow D< 4+\frac{2-1}{1,2}+\frac{3-2}{2.3}+...+\frac{1008-1007}{1007.1008}\)
\(\Leftrightarrow D< 4+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{107}-\frac{1}{1008}\)
\(\Leftrightarrow D< 5-\frac{1}{1008}< 5< 6\)
a: \(A=\dfrac{3^6\cdot3^8\cdot5^4-3^{13}\cdot5^{13}\cdot5^{-9}}{3^{12}\cdot5^6+5^6\cdot3^{12}}\)
\(=\dfrac{3^{14}\cdot5^4-3^{13}\cdot5^4}{2\cdot3^{12}\cdot5^6}\)
\(=\dfrac{3^{13}\cdot5^4\cdot\left(3-1\right)}{2\cdot3^{12}\cdot5^6}=\dfrac{3}{5^2}=\dfrac{3}{25}\)
c: \(C=\dfrac{\dfrac{27}{64}+\dfrac{125}{64}-5\cdot\dfrac{16-15}{12}}{\dfrac{25}{64}+\dfrac{4}{9}-\dfrac{5}{6}}\)
\(=\dfrac{47}{24}:\dfrac{1}{576}=47\cdot24=1128\)
\(D=\dfrac{\left(2!\right)^2}{1^2}+\dfrac{\left(2!\right)^2}{3^2}+\dfrac{\left(2!\right)^2}{5^2}+...+\dfrac{\left(2!\right)^2}{2015^2}\)
\(D=\left(2!\right)^2\left(\dfrac{1}{3^2}+\dfrac{1}{5^2}+...+\dfrac{1}{2015^2}\right)\)
Xét số hạng tổng quát dạng: \(\dfrac{1}{\left(2n+1\right)^2}\) với \(n\in N\ge1\)
Ta có: \(\left(2n+1\right)^2-2n\left(2n+1\right)=1>0\)
\(\Rightarrow\left(2n+1\right)^2>2n\left(2n+1\right)\Rightarrow\dfrac{1}{\left(2n+1\right)^2}< \dfrac{1}{2n\left(2n+1\right)}\)
Do đó: \(\left\{{}\begin{matrix}\dfrac{1}{3^2}< \dfrac{1}{2.4}\\\dfrac{1}{5^2}< \dfrac{1}{4.6}\\....\\\dfrac{1}{2015^2}< \dfrac{1}{2014.2016}\end{matrix}\right.\)
\(\Rightarrow\dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{5^2}...+\dfrac{1}{2015^2}< 1+\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{2014.2016}\)
\(\Leftrightarrow\dfrac{D}{\left(2!\right)^2}< 1+\dfrac{1}{2.4}+\dfrac{1}{4.6}+..+\dfrac{1}{2014.2016}\)
\(\Leftrightarrow D< 4\left(1+\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{2014.2016}\right)\)
\(\Leftrightarrow D< 4+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{1007.1008}\)
\(\Leftrightarrow D< 4+\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+...+\dfrac{1008-1007}{1007.1008}\)
\(\Leftrightarrow D< 4+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{...1}{1007}-\dfrac{1}{1008}\)
\(\Leftrightarrow D< 5-\dfrac{1}{1008}< 5< 6\)
Cám ơn bạn :)