Chứng minh rằng: \(4^{2016}+4^{2017}+4^{2018}⋮84\)
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TT
1
YY
19 tháng 4 2018
\(A=\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2015}\)
\(=\frac{2016-1}{2016}+\frac{2017-1}{2017}+\frac{2018-1}{2018}+\frac{2015+3}{2015}\)
\(=1-\frac{1}{2016}+1-\frac{1}{2017}+1-\frac{1}{2018}+1+\frac{3}{2015}\)
\(=4+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2017}+\frac{1}{2015}-\frac{1}{2018}\)
mà \(\frac{1}{2015}>\frac{1}{2016};\frac{1}{2017};\frac{1}{2018}\)
\(\Rightarrow A>4\)
TP
1
11 tháng 4 2019
\(A=\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2015}\)
\(=\left(1-\frac{1}{2016}\right)+\left(1-\frac{1}{2017}\right)+\left(1-\frac{1}{2018}\right)+\left(1+\frac{3}{2015}\right)\)
\(=1-\frac{1}{2016}+1-\frac{1}{2017}+1-\frac{1}{2018}+1+\frac{1}{2015}+\frac{1}{2015}+\frac{1}{2015}\)
\(=\left(1+1+1+1\right)+\left(\left(\frac{1}{2015}-\frac{1}{2016}\right)+\left(\frac{1}{2015}-\frac{1}{2017}\right)+\left(\frac{1}{2015}-\frac{1}{2018}\right)\right)\)
\(=4+\left(\frac{1}{2015}-\frac{1}{2016}\right)+\left(\frac{1}{2015}-\frac{1}{2017}\right)+\left(\frac{1}{2015}-\frac{1}{2018}\right)\)
Vì \(\frac{1}{2015}>\frac{1}{2016};\frac{1}{2015}>\frac{1}{2017};\frac{1}{2015}>\frac{1}{2018}\)
\(\Rightarrow\frac{1}{2015}-\frac{1}{2016}>0;\frac{1}{2015}-\frac{1}{2017}>0;\frac{1}{2015}-\frac{1}{2018}>0\)
\(\Rightarrow\left(\frac{1}{2015}-\frac{1}{2016}\right)+\left(\frac{1}{2015}-\frac{1}{2017}\right)+\left(\frac{1}{2015}-\frac{1}{2018}\right)>0\)
\(\Rightarrow4+\left(\frac{1}{2015}-\frac{1}{2016}\right)+\left(\frac{1}{2015}-\frac{1}{2017}\right)+\left(\frac{1}{2015}-\frac{1}{2018}\right)>4\)
\(\Rightarrow A>4\left(dpcm\right)\)
LG
0
NH
1
ML
1
19 tháng 10 2022
Đặt x/3=y/4=z/5=k
=>x=3k; y=4k; z=5k
\(\dfrac{4x-3y}{2016}=\dfrac{4\cdot3k-3\cdot4k}{2016}=0\)
\(\dfrac{5y-4z}{2017}=\dfrac{5\cdot4k-4\cdot5k}{2017}=0\)
\(\dfrac{3z-5x}{2018}=\dfrac{3\cdot5k-5\cdot3k}{2018}=0\)
=>\(\dfrac{4x-3y}{2016}=\dfrac{5y-4z}{2017}=\dfrac{3z-5x}{2018}\)
24 tháng 4 2019
\(A=\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2015}\)
\(A=1-\frac{1}{2016}+1-\frac{1}{2017}+1-\frac{1}{2018}+1+\frac{3}{2015}\)
\(A=4-\left(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}-\frac{3}{2015}\right)\)
Xét :
\(\frac{1}{2016}< \frac{1}{2015}\)\(;\)\(\frac{1}{2017}< \frac{1}{2015}\)\(;\)\(\frac{1}{2018}< \frac{1}{2015}\)
\(\Rightarrow\)\(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}< \frac{1}{2015}+\frac{1}{2015}+\frac{1}{2015}\)
\(\Leftrightarrow\)\(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}-\frac{3}{2015}< 0\)
Suy ra : \(A=4-\left(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}-\frac{3}{2015}\right)>4-0=4\) ( đpcm )
...
\(4^{2016}+4^{2017}+4^{2018}\)
\(=4^{2015}\cdot\left(4+4^2+4^3\right)\)
\(=4^{2015}\cdot84⋮84\left(đpcm\right)\)