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)}

Bài 4:

\(a,\Rightarrow\left(x+2\right)\left(y+1\right)=3\cdot7=7\cdot3=21\cdot1=1\cdot21\)

Vậy \(\left(x;y\right)\in\left\{\left(19;0\right);\left(1;6\right);\left(5;2\right)\right\}\)

b: 4/x+y/3=5/6

=>\(\dfrac{12+xy}{3x}=\dfrac{5}{6}=\dfrac{5x}{6x}\)

=>24+2xy=5x

=>5x-2xy=24

=>x(5-2y)=24

=>x(2y-5)=-24

=>(x;2y-5) thuộc {(24;-1); (-24;1); (8;-3); (-8;3)}(Vì x và y là số nguyên)

=>(x,y) thuộc {(24;2); (-24;3); (8;1); (-8;1)}

a: =>\(\dfrac{xy-12}{3y}=\dfrac{1}{5}\)

=>5(xy-12)=3y

=>5xy-3y=60

=>y(5x-3)=60

=>(y;5x-3) thuộc {(5;12); (30;2)}(Vì x,y là số nguyên)

=>(y,x) thuộc {(5;3); (30;1)}

b: Bạn ghi lại đề đi bạn

Câu 2: 

a: \(\Leftrightarrow x+2\in\left\{3;9\right\}\)

hay \(x\in\left\{1;7\right\}\)

Thi à :)?

a: \(\left(x,y\right)\in\left\{\left(1;2\right);\left(-1;-2\right);\left(2;1\right);\left(-2;-1\right)\right\}\)

\(a,\Leftrightarrow7⋮x-1\Leftrightarrow x-1\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\\ \Leftrightarrow x\in\left\{-6;0;2;8\right\}\\ b,\Leftrightarrow\dfrac{x-1+2}{x-1}\in Z\Leftrightarrow1+\dfrac{2}{x-1}\in Z\\ \Leftrightarrow2⋮x-1\Leftrightarrow x-1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\\ \Leftrightarrow x\in\left\{-1;0;2;3\right\}\)

a) để 7/x-1 thuộc Z 

=> (x-1) thuộc ước 7(+-1;+-7)

x-1  -1     1      -7      7

x      0     2       -6     8

a) \(\dfrac{7}{x-1}\in Z\Leftrightarrow x-1\inƯ\left(7\right)=\left\{1;-1;7;-7\right\}\)

\(\Rightarrow x\in\left\{2;0;8;-6\right\}\)

b) \(\dfrac{x+1}{x-1}\in Z\Rightarrow\dfrac{x-1}{x-1}+\dfrac{2}{x-1}=1+\dfrac{2}{x-1}\in Z\)

\(\Rightarrow x-1\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)

\(\Rightarrow x\in\left\{2;0;3;-1\right\}\)

Bài 1: 

c: x=4

b: x=2

a) (x - 140) : 7 = 3³ - 2³ . 3

(x - 140) : 7 = 27 - 8 . 3 = 27 - 24 = 3

x - 140 = 3 x 7 = 21

x = 21 + 140 = 161

b) x³ . x² = 2⁸ : 2³

x⁵ = 2⁵

=> x = 2

c) (x + 2) . ( x - 4) = 0

x = -2 hoặc 4

d) 3^x-3 - 3² = 2 . 3² =

3^x-3 - 9 = 2 . 9 = 18

3^x-3 = 18 + 9 = 27

3^x-3 = 3³

=> x - 3 = 3

x = 3 + 3 = 6

2.

a) 9 : ( x + 2 )

9 ⋮ 1 ; 9 ⋮ 3 ; 9 ⋮ 9

=> x = -1 ; 1 ; 7

a) Để y nguyên thì \(6x-4⋮2x+3\)

\(\Leftrightarrow-13⋮2x+3\)

\(\Leftrightarrow2x+3\in\left\{1;-1;13;-13\right\}\)

\(\Leftrightarrow2x\in\left\{-2;-4;10;-16\right\}\)

hay \(x\in\left\{-1;-2;5;-8\right\}\)