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Để \(11 \vdots 2x-1\) thì \(2x-1 \in Ư_{11}\)
Mà `Ư_{11}=`{\(\pm 1;\pm 11\)}
`@2x-1=1=>2x=2=>x=1`
`@2x-1=-1=>2x=0=>x=0`
`@2x-1=11=>2x=12=>x=6`
`@2x-1=-11=>2x=-10=>x=-5`
11⋮ 2x-1
=> 2x-1 ϵ U(11)= (1;-1;11;-11)
- 2x-1=1
2x=2
x=2:2
x=1
- 2x-1=-1
2x =-1+1
2x =0
x =0
- 2x-1 =11
2x =11+1
2x = 12
x=6
- 2x-1=-11
2x =-11+1
2x =-10
x=-5
x ϵ (1;0;6;-5)
nếu đúng thì tích đúng cho mình với nha
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3a+11=\(3a+6+5=3\left(a+1\right)+5\)
Để 3a+11 \(⋮\left(a+2\right)\Leftrightarrow3\left(a+2\right)+5⋮\left(a+2\right)\Leftrightarrow5⋮\left(a+2\right)\Rightarrow a+2\in\left\{-5;-1;1;5\right\}\Rightarrow a\in\left\{-7;-3;-1;3\right\}\)
Ta có: \(3a+11⋮a+2\)
<=> \(3\left(a+2\right)+5⋮a+2\)
<=> \(5⋮a+2\)
Vì a nguyên => \(a+2\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
Ta có bảng sau:
a+2 | 1 | -1 | 5 | -5 |
a | -1 | -3 | 3 | -7 |
Vậy \(a\in\left\{-1;-3;3;-7\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(3a+11\right)⋮\left(a+2\right)\\ \Rightarrow\left[3\left(a+2\right)+5\right]⋮\left(a+2\right)\\ \Rightarrow5⋮\left(a+2\right)\left[vì:3\left(a+2\right)⋮\left(a+2\right)\right]\\ \Rightarrow\left(a+2\right)\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\\ \left(a+2\right)\in\left\{-1;-3;3;-7\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
S=(1+2+2^2+2^3+2^4+2^5)+(2^6+2^7+2^8+2^9+2^10+2^11)
S=63.2^6(1+2+2^2+2^3+2^4+2^5)
S=63+2^6.63
S=63.(1+2^6)
S=63.65:hết cho 9
\(\Rightarrow\) đpcm