CMR: \(x^{2017}+x^{2015}+1⋮x^2+x+1\)
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2 tháng 9 2019
=> (x+2020)/5=(x+2020)/6=(x+2020)/3+(x+2020)/2
=>(x+2020)(1/5+1/6)=(x+2020)(1/3+1/2)
Với x+2020=0=>x=-2020
Với x+2020 khác 0=>1/5+1/6=1/3+1/2 ,vô lí
Vậy x=-2020
SK
3
27 tháng 8 2021
\(x-\dfrac{1}{2017}+x-\dfrac{2}{2016}=x-\dfrac{3}{2015}+x-\dfrac{4}{2014}\)
\(\Rightarrow x+x-x-x=-\dfrac{4}{2014}-\dfrac{3}{2015}+\dfrac{2}{2016}+\dfrac{1}{2017}\)
\(\Rightarrow0=-\dfrac{4}{2014}-\dfrac{3}{2015}+\dfrac{2}{2016}+\dfrac{1}{2017}\) (vô lí)
27 tháng 8 2021
\(\dfrac{x-1}{2017}+\dfrac{x-2}{2016}=\dfrac{x-3}{2015}+\dfrac{x-4}{2014}\Leftrightarrow\left(\dfrac{x-1}{2017}-1\right)+\left(\dfrac{x-2}{2016}-1\right)=\left(\dfrac{x-3}{2015}-1\right)+\left(\dfrac{x-4}{2014}-1\right)\Leftrightarrow\left(x-2018\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\Leftrightarrow x=2018\)
22 tháng 3 2023
\(\dfrac{x-1}{2021}+\dfrac{x-2}{2020}=\dfrac{x-5}{2017}+\dfrac{x-7}{2015}\\ \dfrac{x-1}{2021}+\dfrac{x-2}{2020}-2=\dfrac{x-5}{2017}+\dfrac{x-7}{2015}-2\\ \dfrac{x-1}{2021}+\dfrac{x-2}{2020}-1-1=\dfrac{x-5}{2017}+\dfrac{x-7}{2015}-1-1\\\left(\dfrac{x-1}{2021}-1\right)+\left(\dfrac{x-2}{2020}-1\right)=\left(\dfrac{x-5}{2017}-1\right)+\left(\dfrac{x-7}{2015}-1\right)\\ \dfrac{x-2022}{2021}+\dfrac{x-2022}{2020}=\dfrac{x-2022}{2017}+\dfrac{x-2022}{2015}\\ \dfrac{x-2022}{2021}+\dfrac{x-2022}{2020}-\dfrac{x-2022}{2017}-\dfrac{x-2022}{2015}=0\\ \left(x-2022\right)\left(\dfrac{1}{2021}+\dfrac{1}{2020}-\dfrac{1}{2017}-\dfrac{1}{2015}\right)=0\)
mà `(1/2021+1/2020-1/2017-1/2015 \ne 0`
nên `x-2022=0`
`x=2022`
G
0
23 tháng 6 2015
1) 9 x 10 + 10 x 11 + 11 x 12 + ....+ 2015 x 2016
=9x10x11-8x9x10+10x11x12-9x10x11+...+2015x2016x2017-2014x2015x2013
=2015x2016x2017-8x9x10
=8193537360
ta có x2017 + x2015 + 1 = x2015.( x2 + x + 1)
do x2 + x + 1 chia hết cho x2 + x + 1
nên x2017 + x2015 + 1 chia hết cho x2+x+1