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![](https://rs.olm.vn/images/avt/0.png?1311)
1/
$C=5+(5^2+5^3)+(5^4+5^5)+.....+(5^{2022}+5^{2023})$
$=5+5^2(1+5)+5^4(1+5)+....+5^{2022}(1+5)$
$=5+(1+5)(5^2+5^4+....+5^{2022})$
$=5+6(5^2+5^4+....+5^{2022})$
$\Rightarrow C$ chia $6$ dư $5$
$\Rightarrow C\not\vdots 6$
2/
$D=(1+2+2^2)+(2^3+2^4+2^5)+....+(2^{2019}+2^{2020}+2^{2021})$
$=(1+2+2^2)+2^3(1+2+2^2)+....+2^{2019}(1+2+2^2)$
$=(1+2+2^2)(1+2^3+...+2^{2019})$
$=7(1+2^3+...+2^{2019})\vdots 7$
Ta có đpcm.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 6:
\(\Leftrightarrow6n+4⋮2n-1\)
\(\Leftrightarrow2n-1\in\left\{1;-1;7;-7\right\}\)
hay \(n\in\left\{1;0;4;-3\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Ư(8)={1;2;4;8}
Ư(12)={1;2;3;4;6;12}
UC(8;12)={1;2;4}
b: B(16)={0;16;32;...}
B(24)={0;24;48;...}
BC(16,24)={0;48;96;...}
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{-13}{8}+\dfrac{-5}{9}+\dfrac{26}{26}-\dfrac{13}{9}\)
= \(\left(\dfrac{-13}{8}+\dfrac{26}{16}\right)+\left(\dfrac{-5}{9}-\dfrac{13}{9}\right)\)
= \(\left(\dfrac{-26}{16}+\dfrac{26}{26}\right)+\left(\dfrac{-18}{9}\right)\)
= \(0+\left(-2\right)\)
= \(-2\)
\(\left(\dfrac{-13}{8}-\dfrac{26}{16}\right)+\left(\dfrac{-5}{9}-\dfrac{13}{9}\right)=\left(\dfrac{-13}{8}-\dfrac{13}{8}\right)+\dfrac{-18}{9}=0+\left(-2\right)=-2\)