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Ta có : \(2^{2015}+3^{2017}=2^{2012+3}+3^{2016+1}\) = \(2^{2012}.2^3+3^{2016}.3\)
= \(2^{2012}.8+3^{2016}.3\) = (........6).8 + (.......1) .3 = (.........8) + (..........3) = (........1)
=> \(2^{2015}+3^{2017}\) chia 5 dư 1.
k nha bạn
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Ta có \(B=\frac{2015+2016+2017}{2016+2017+2018}\)
\(\Leftrightarrow B=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
Vì
\(\frac{2015}{2016}>\frac{2015}{2016+2017+2018};\frac{2016}{2017}>\frac{2016}{2016+2017+2018};\frac{2017}{2018}>\frac{2017}{2016+2017+2018}\) nên \(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
Hay \(A>B\)
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dư 1 hoặc 0 vì
nếu 12 thì 1 : 3 = 0 dư 1
22 thì 4 : 3 = 1 dư 1
32 thì 9 : 3 = 3
42 thì 16 : 3 = 5 dư 1
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a) Trong phép chia cho 3 số dư có thể là 0, 1, 2
________________ 4 _________________, 3
________________ 5 ___________________4
b) Số chia hết vcho 3 là 3k, chia 3 dư 1 là 3k+1, chia 3 dư 2 là 3k+2
Giúp mình với mình đang cần gấp !!!