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a)2017^2018 có tận cùng là 9
b)Khi 2017^2018 chia cho 5 thì dư 4
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{2017^{2018+1}}{2017^{2018-3}}\)và \(B=\frac{2017^{2018-1}}{2017^{2018-5}}\)
Có \(A=\frac{2017^{2019}}{2017^{2015}}\)và \(B=\frac{2017^{2017}}{2017^{2013}}\)
Mà\(\frac{2017^{2019}}{2017^{2015}}>\frac{2017^{2018}}{2017^{2015}}\)và\(\frac{2017^{2017}}{2017^{2013}}>\frac{2017^{2017}}{2017^{2015}}\)
Vì \(\frac{2017^{2018}}{2017^{2015}}>\frac{2017^{2017}}{2017^{2015}}\)
Vậy A>B
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\(\left|2x-28\right|^{2018}+\left(3y+9\right)^{2020}=0\)
Giải:Vì \(\left|2x-28\right|^{2018}\ge0\);\(\left(3y+9\right)^{2020}\ge0\)
\(\Rightarrow\left|2x-28\right|^{2018}+\left(3y+9\right)^{2020}\ge0\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}\left|2x-28\right|=0\\3y+9=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}2x-28=0\\3y+9=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=14\\y=-3\end{cases}}\)
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Bài 1:
|x-2|=4-x
ĐK: \(4-x\ge0\Leftrightarrow x\le4\)
Ta có: \(\orbr{\begin{cases}x-2=4-x\\x-2=x-4\end{cases}\Rightarrow\orbr{\begin{cases}2x=6\\0=2\left(loại\right)\end{cases}\Rightarrow}}x=3\left(tm\right)\)
Vậy x = 3
Bài 2:
a, sao có z
b, Vì \(\hept{\begin{cases}\left|2017-x\right|\ge0\\\left|y-x+2018\right|\ge0\end{cases}\Rightarrow\left|2017-x\right|+\left|y-x+2018\right|\ge0}\)
Mà |2017-x|+|y-x+2018|=0
\(\Rightarrow\hept{\begin{cases}\left|2017-x\right|=0\\\left|y-x+2018\right|=0\end{cases}\Rightarrow\hept{\begin{cases}x=2017\\y-2017+2018=0\end{cases}\Rightarrow}\hept{\begin{cases}x=2017\\y=1\end{cases}}}\)
Vậy x=2017,y=1
c, giống b
1
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