\(a+2⋮a-1\)

\(=>\left(a-1\right)+3⋮a-1\)

\(\)Vì \(a-1⋮a-1\) mà \(\left(a-1\right)+3⋮a-1\)

\(=>3⋮a-1\)

\(=>a\in\text{Ư}\left(3\right)=\left\{-3;-1;1;3\right\}\)

co a+2=a-1+3

de a+2 chia het cho a-1 thi 3 chia het cho a-1

=> a-1 thuoc uoc cua 3

ma U(3)∈{-1;1;-3;3}

ta co bang sau

vay...

\(\left(a+2\right)⋮\left(a-1\right)\)

\(\left(a-1+3\right)⋮\left(a-1\right)\) 

\(\text{ }\Rightarrow a-1\in\text{Ư}\left(3\right)=\left\{\pm1;\pm3\right\}\)

`+, a-1=1 => a=2`

`+,a-1=-1=>a=0`

`+, a-1=3=>x=4`

`+,a-1=-3=>a=-2`

vậy \(a\in\left\{2;0;4;-2\right\}\)

ta có a+5=a+5+3

=> 3 chia hết cho a+5

a nguyên => a+5 nguyên 

=> a+5\(\in\)Ư(3)={-3;-1;1;3}

ta có bảng

vậy a={-8;-6;-4;-2}

Ta có: \(a+8⋮a+5\)

\(\Leftrightarrow a+5+3⋮a+5\)

\(\Leftrightarrow3⋮a+5\)

\(\Rightarrow a+5\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

Nếu a + 5 = -1 => a = -6

a + 5 = 1 => a = -4

a + 5 = 3 => a = -2

a + 5 = -3 => a = -8

Vậy \(a=\left\{-6;-4;-2;-8\right\}\)thì \(a+8⋮a+5\)

a^2-2 < 0

=> a^2 < 2

Mà a^2 >= 0

=> 0 < = a^2 < 2

=> a^2 thuộc {0;1}

=> a thuộc {-1;0;1}

Vậy .......

Tk mk nha

a se la cac so nguyen am

a: \(\Leftrightarrow\left(a^2-3\right)\left(a^2-10\right)< 0\)

\(\Leftrightarrow3< a^2< 10\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{3}< a< \sqrt{10}\\-\sqrt{10}< a< -\sqrt{3}\end{matrix}\right.\)

d: \(\Leftrightarrow\left(a-3\right)\left(a+1\right)< 0\)

hay -1<a<3

a: =>7/4<a<80

=>\(a\in\left\{2;3;...;79\right\}\)

b: =>19/10<a<130

=>\(a\in\left\{2;3;...;129\right\}\)

c: =>32<a<490

=>\(a\in\left\{33;34;...;489\right\}\)

2:

a: 5/x-y/3=1/6

=>\(\dfrac{15-xy}{3x}=\dfrac{1}{6}\)

=>\(\dfrac{30-2xy}{6x}=\dfrac{x}{6x}\)

=>30-2xy=x

=>x(2y+1)=30

=>(x;2y+1) thuộc {(30;1); (-30;-1); (10;3); (-10;-3); (6;5); (-6;-5)}

=>(x,y) thuộc {(30;0); (-30;-1); (10;1); (-10;-2); (6;2); (-6;-3)}

b: x/6-2/y=1/30

=>\(\dfrac{xy-12}{6y}=\dfrac{1}{30}\)

=>\(\dfrac{5xy-60}{30y}=\dfrac{y}{30y}\)

=>5xy-60=y

=>y(5x-1)=60

=>(5x-1;y) thuộc {(-1;-60); (4;15); (-6;-10)}(Vì x,y là số nguyên)

=>(x,y) thuộc {(0;-60); (1;15); (-1;-10)}