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\(2015^{2017}+2017^{2015}=\left(2015^{2017}+1\right)+\left(2017^{2015}-1\right)=A\left(2015+1\right)+B\left(2017-1\right)=2016A+2016B=2016\left(A+B\right)\)Luôn chia hết cho 2016
Vậy ta có điều phải chứng minh.
Lời giải:
Ta có:
\(A=2015^{2017}+2017^{2015}=2015^{2017}+1+2017^{2015}-1\)
Theo khai triển hằng đẳng thức:
\(2015^{2017}+1=2015^{2017}+1^{2017}=(2015+1)(2015^{2016}-2015^{2015}+....-2015+1)\vdots (2015+1)\)
\(\Leftrightarrow 2015^{2017}+1\vdots 2016\) (1)
Và: \(2017^{2015}-1=2017^{2015}-1^{2015}=(2017-1)(2017^{2014}+2017^{2013}+...+2017+1)\vdots (2017-1)\)
\(\Leftrightarrow 2017^{2015}-1\vdots 2016\) (2)
Từ (1),(2) suy ra \(A=2015^{2017}+2017^{2015}\vdots 2016\) (đpcm)
Nếu đúng tick em nha
2015^2017+2017^2015
=2015^2017+2017^2015-1
=(2015^2017+1^2017)+(2017^2015-1^2015)
Do 2015^2017+1^2017\(⋮\)2015+2=2016
2017^2015-1^2015\(⋮\)2017-1=2016
Vậy (2015^2017+2017^2015)\(⋮\)2016
Tick nha !
\(\frac{x+5}{2015}+\frac{x+4}{2016}+\frac{x+3}{2017}=\frac{x+2015}{5}+\frac{x+2016}{4}+\frac{x+2017}{3}\)
\(\Leftrightarrow\frac{x+5}{2015}+\frac{x+4}{2016}+\frac{x+3}{2017}-\frac{x+2015}{5}-\frac{x+2016}{4}-\frac{x+2017}{3}=0\)
\(\Leftrightarrow\left(\frac{x+5}{2015}+1\right)+\left(\frac{x+4}{2016}+1\right)+\left(\frac{x+3}{2017}+1\right)-\left(\frac{x+2015}{5}+1\right)-\left(\frac{x+2016}{4}+1\right)\)
\(-\left(\frac{x+2017}{3}+1\right)=0\)
\(\Leftrightarrow\frac{x+2020}{2015}+\frac{x+2020}{2016}+\frac{x+2020}{2017}-\frac{x+2020}{5}-\frac{x+2020}{4}-\frac{x+2020}{3}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\right)=0\)
\(\Leftrightarrow x+2020=0\left(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\ne0\right)\)
<=> x=-2020
Vậy x=-2020
Chưa ai giải thì thui
MK cũng bó tay
Chúc bn hok t
:) hihi