4^2017:(4^2014+3x4^2014)
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\(^{4^{2017}:\left(4^{2014}\left(1+3\right)\right)}\)
\(4^{2017}:4^{2015}=4^2=16\)
\(4^{2017}:\left(4^{2014}.\left(1+3\right)\right)\)
\(=4^{2017}:4^{2014}.5\)
\(=64×5\)
\(=320\)
42017 : ( 42014 + 3 . 42014 )
= 42017 : [ 42014 ( 1 + 3 ) ]
= 42017 : [ 42014 . 4 ]
= 42017 : 42015
= 42 = 16
=))
\(4^{2017}:\left(4^{2014}+3\cdot4^{2014}\right)\)
\(=4^{2017}:4^{2014}\left(1+3\right)\)
\(=4^3\cdot4\)
\(=4^4\)
\(=256\)
Cho A= \(\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2014}.\). So sánh A với 4
\(A=\dfrac{2014}{2015}+\dfrac{2015}{2016}+\dfrac{2016}{2017}+\dfrac{2017}{2014}\\ =1-\dfrac{1}{2015}+1-\dfrac{1}{2016}+1-\dfrac{1}{2017}+1+\dfrac{1}{2014}+\dfrac{1}{2014}+\dfrac{1}{2014}\\ =\left(1+1+1+1\right)+\left[-\left(\dfrac{1}{2015}-\dfrac{1}{2014}+\dfrac{1}{2016}-\dfrac{1}{2014}+\dfrac{1}{2017}-\dfrac{1}{2014}\right)\right]\\ =4+\left[-\left(\dfrac{1}{2015}-\dfrac{1}{2014}+\dfrac{1}{2016}-\dfrac{1}{2014}+\dfrac{1}{2017}-\dfrac{1}{2014}\right)\right]\)
Vì \(\dfrac{1}{2015}< \dfrac{1}{2014}\), \(\dfrac{1}{2016}< \dfrac{1}{2014}\), \(\dfrac{1}{2017}< \dfrac{1}{2014}\)
\(\Rightarrow\left(\dfrac{1}{2015}-\dfrac{1}{2014}+\dfrac{1}{2016}-\dfrac{1}{2014}+\dfrac{1}{2017}-\dfrac{1}{2014}\right)< 0\\ \Rightarrow-\left(\dfrac{1}{2015}-\dfrac{1}{2014}+\dfrac{1}{2016}-\dfrac{1}{2014}+\dfrac{1}{2017}-\dfrac{1}{2014}\right)\\>0\\ \Rightarrow4+\left[-\left(\dfrac{1}{2015}-\dfrac{1}{2014}+\dfrac{1}{2016}-\dfrac{1}{2014}+\dfrac{1}{2017}-\dfrac{1}{2014}\right)\right]>4\)
số số hạng là:
(2018-2):1+1=2017( số hạng)
[2+(-3)]+[4+(-5)+...+[2016+(-2017)]+2018
=(-1)+(-1)+...+(-1)+2018
=(-1008)+2018
=1010