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![](https://rs.olm.vn/images/avt/0.png?1311)
bài 2:để Z là số nguyên thì 3n-5 \(⋮\)n+4
\(\Rightarrow[(3n-5)-3(n+4)]⋮(n+4)\)
\(\Rightarrow(3n-5-3n-12)⋮(n+4)\)
\(\Rightarrow-17⋮n+4\)
\(\Rightarrow n+4\inƯ(17)\)={1;-1;17;-17}
\(\Rightarrow\)n\(\in\){-3;-5;13;-21}
![](https://rs.olm.vn/images/avt/0.png?1311)
4,
a,\(\dfrac{x-1}{9}\)=\(\dfrac{8}{3}\)
[x- 1].3=9.8
[x- 1].3=72
x-1=72:3
x-1=24
x=24+1
x=25
![](https://rs.olm.vn/images/avt/0.png?1311)
a, Ta có
\(\dfrac{111}{37}< x< \dfrac{91}{37}\Leftrightarrow3< x< 7\Leftrightarrow x\in\left\{4,5,6\right\}\)
b,Ta co
\(\dfrac{-84}{14}< 3x< \dfrac{108}{9}\Leftrightarrow3.\dfrac{-28}{14}< 3x< 3.\dfrac{36}{9}\Leftrightarrow\dfrac{-28}{14}< x< \dfrac{36}{9}\Leftrightarrow-2< x< 4\Leftrightarrow x\in\left\{-1,0,1,2,3\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Theo đề bài 3<x<7.
=> x\(\in\){4;5;6}
b) tương tự như a)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 111/37 =3
91/13=7
-> x \(\in\left\{4,5,6\right\}\)
b) 84/14=6
108/9=12
mà 6<3x<12
-> 3x=9
->x=3.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
`a)(4,5-2x)*1 4/7=11/14`
`=>(4,5-2x)*11/7=11/14`
`=>4,5-2x=1/2`
`=>2x=4,5-0,5=4`
`=>x=2`
Vậy `x=2`
`b)(2,8x-32):2/3=-90`
`=>2,8x-32=-90*2/3=-60`
`=>2,8x=-28`
`=>x=-10`
Vậy `x=-10`
a)
\(\dfrac{111}{37}=3< x< \dfrac{91}{13}=7\)
Vậy x = {4;5;6}
b)
\(-\dfrac{84}{14}=-6< 3x< \dfrac{108}{9}=12\Leftrightarrow-2< x< 4\)
Vậy x = {-1;0;1;2;3}
a, Ta có : \(\dfrac{111}{37}< x< \dfrac{91}{13}\)
\(\Rightarrow3< x< 7\)
Mà x là số nguyên .
\(\Rightarrow x\in\left\{4;5;6\right\}\)
b, Ta có : \(-\dfrac{84}{14.3}< x< \dfrac{108}{9.3}\)
\(\Rightarrow-2< x< 4\)
Mà x là số nguyên .
\(\Rightarrow x\in\left\{-1;0;1;2;3\right\}\)