\(A=\left|2x-4\right|+\left(y+10\right)^{2016}+2016>=2016\)
Dấu '=' xảy ra khi x=2 và y=-10
`a)`\(A=2016+\left|2x-4\right|+\left(y+10\right)^{2016}\)
Ta có: \(\left\{{}\begin{matrix}\left|2x-4\right|\ge0\\\left(y+10\right)^{2016}\ge0\end{matrix}\right.\)
\(\Rightarrow A\ge2016\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}2x-4=0\\y+10=0\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}x=2\\y=-10\end{matrix}\right.\)
Vậy \(Min_A=2016\) khi \(\left\{{}\begin{matrix}x=2\\y=-10\end{matrix}\right.\)
`b)`\(A=\dfrac{1.4}{2.3}+\dfrac{2.5}{3.4}+\dfrac{3.6}{4.5}+...+\dfrac{2013.2016}{2014.2015}\)
Ta có công thức tổng quát: \(\dfrac{n\left(n+3\right)}{\left(n+1\right)\left(n+2\right)}=\dfrac{n^2+3n+2-2}{n^2+3n+2}\)
\(=1-\dfrac{2}{n^2+3n+2}\)
\(=1-\dfrac{2}{\left(n+1\right)\left(n+2\right)}\)
\(\Rightarrow\dfrac{1.4}{2.3}=1-\dfrac{2}{2.3}\)
\(\dfrac{2.5}{3.4}=1-\dfrac{2}{3.4}\)
\(\dfrac{3.6}{4.5}=1-\dfrac{2}{4.5}\)
\(\rightarrow A=2013-2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{2014.2015}\right)\)
\(A=2013-2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{2014}-\dfrac{1}{2015}\right)\)
\(A=2013-2\left(\dfrac{1}{2}-\dfrac{1}{2015}\right)\)
\(A=2013-1+\dfrac{2}{2015}\)
\(A=2012+\dfrac{2}{2015}\)
\(\Rightarrow2012< A< 2013\left(đfcm\right)\)
