![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\Rightarrow\left(n+2\right)+3⋮\left(n+2\right)\)
\(\Rightarrow\left(n+2\right)\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)
\(\Rightarrow n\in\left\{-5;-3;-1;1\right\}\)
b) \(\Rightarrow\left(n+1\right)+6⋮\left(n+1\right)\)
\(\Rightarrow\left(n+1\right)\inƯ\left(6\right)=\left\{-6;-3;-2;-1;1;2;3;6\right\}\)
\(\Rightarrow n\in\left\{-7;-4;-3;-2;0;1;2;5\right\}\)
c) \(\Rightarrow\left(n+1\right)^2-\left(n+1\right)+13⋮\left(n+1\right)\)
\(\Rightarrow\left(n+1\right)\inƯ\left(13\right)=\left\{-13;-1;1;13\right\}\)
\(\Rightarrow n\in\left\{-14;-2;0;12\right\}\)
d) \(\Rightarrow\left(n+2\right)^2-\left(n+2\right)+7⋮\left(n+2\right)\)
\(\Rightarrow\left(n+2\right)\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\)
\(\Rightarrow n\in\left\{-9;-3;-1;5\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\) \(ƯCLN\left(a,b\right)=18\Rightarrow\left\{{}\begin{matrix}a=18m\\b=18n\end{matrix}\right.;\left(m,n\right)=1\)
Thay vào \(a+b=162\)
\(\Rightarrow18\left(m+n\right)=162\\ \Rightarrow m+n=9\)
Vì \(\left(m,n\right)=1\)
m | n | a | b |
1 | 8 | 18 | 144 |
2 | 7 | 36 | 126 |
4 | 5 | 72 | 90 |
5 | 4 | 90 | 72 |
7 | 2 | 126 | 36 |
8 | 1 | 144 | 18 |
Vậy \(\left(a;b\right)=\left\{\left(18;144\right);\left(36;126\right);\left(72;90\right);\left(90;72\right);\left(126;36\right)\left(144;18\right)\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(3x-4\right)^3=5^2+4.5^2\)
\(\Leftrightarrow\left(3x-4\right)^3=5^2\left(1+4\right)\)
\(\Leftrightarrow\left(3x-4\right)^3=5^3\)
\(\Leftrightarrow3x-4=5\Leftrightarrow3x=9\Leftrightarrow x=3\)
Ta có: \(\left(3x-4\right)^3=5^2+4\cdot5^2\)
\(\Leftrightarrow3x-4=5\)
hay x=3
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$A=7+(7^2+7^3+7^4+7^5)+(7^6+7^6+7^8+7^9)+....+(7^{2018}+7^{2019}+7^{2020}+7^{2021})$
$=7+7^2(1+7+7^2+7^3)+7^6(1+7+7^2+7^3)+....+7^{2018}(1+7+7^2+7^3)$
$=7+(1+7+7^2+7^3)(7^2+7^6+....+7^{2018}$
$=7+400(7^2+7^6+....+7^{2018})$
Dễ thấy $400(7^2+7^6+....+7^{2018})$ tận cùng là $0$
Do đó $A$ tận cùng là $7$
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3\left(x+2\right)^3-1^{2019}=5\cdot4^2\)
\(\Leftrightarrow3\left(x+2\right)^3=5\cdot16+1=81\)
\(\Leftrightarrow x+2=3\)
hay x=1
\(A=-a+b+c-d-c+b+a+a-b+d=a+b=1\\ B=-a-b+c+2a+b-3c+2c+3b-4a-a-7b=-4a-4b=-4\left(a+b\right)=-4\)