a) 

\(\left(x+1\right)\left(y-2\right)=5\\ \Rightarrow\left(x+1\right),\left(y-2\right)\inƯ\left(5\right)=\left\{1;-1;5;-5\right\}\)

Ta có bảng:

Vậy \(\left(x;y\right)=\left(0;7\right),\left(-2;-3\right),\left(4;3\right),\left(-6;1\right)\)

b) 

\(\left(x-5\right)\left(y+4\right)=-7\\ \Rightarrow\left(x-5\right),\left(y+4\right)\inƯ\left(-7\right)=\left\{1;-1;7;-7\right\}\)

Ta có bảng:

Vậy \(\left(x;y\right)=\left(6;-11\right),\left(4;3\right),\left(12;-5\right),\left(-2;-3\right)\)

b: =>3x+9=0 và y^2-9=0 và x+y=0

=>x=-3; y=3

a: (2x-5)(y+3)=-22

mà x,y là số nguyên

nên \(\left(2x-5;y+3\right)\in\left\{\left(1;-22\right);\left(11;-2\right);\left(-1;22\right);\left(-11;2\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(3;-25\right);\left(8;-5\right);\left(2;19\right);\left(-3;-1\right)\right\}\)

a: x/2=-5/y

=>xy=-10

=>\(\left(x,y\right)\in\left\{\left(1;-10\right);\left(-10;1\right);\left(-1;10\right);\left(10;-1\right);\left(2;-5\right);\left(-5;2\right);\left(-2;5\right);\left(5;-2\right)\right\}\)

b: =>xy=12

mà x>y>0

nên \(\left(x,y\right)\in\left\{\left(12;1\right);\left(6;2\right);\left(4;3\right)\right\}\)

c: =>(x-1)(y+1)=3

=>\(\left(x-1;y+1\right)\in\left\{\left(1;3\right);\left(3;1\right);\left(-1;-3\right);\left(-3;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(2;2\right);\left(4;0\right);\left(0;-4\right);\left(-2;-2\right)\right\}\)

d: =>y(x+2)=5

=>\(\left(x+2;y\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(-1;5\right);\left(3;1\right);\left(-3;-5\right);\left(-7;-1\right)\right\}\)

a,\(\dfrac{6}{2x+1}=\dfrac{2}{7}\)

⇒\(\dfrac{6}{2x+1}=\dfrac{6}{21}\)

⇒\(2x+1=21\)

\(2x=21-1\)

\(2x=20\)

⇒\(x=10\)

Bài 10:

a: 2x-3 là bội của x+1

=>\(2x-3⋮x+1\)

=>\(2x+2-5⋮x+1\)

=>\(-5⋮x+1\)

=>\(x+1\in\left\{1;-1;5;-5\right\}\)

=>\(x\in\left\{0;-2;4;-6\right\}\)

b: x-2 là ước của 3x-2

=>\(3x-2⋮x-2\)

=>\(3x-6+4⋮x-2\)

=>\(4⋮x-2\)

=>\(x-2\inƯ\left(4\right)\)

=>\(x-2\in\left\{1;-1;2;-2;4;-4\right\}\)

=>\(x\in\left\{3;1;4;0;6;-2\right\}\)

Bài 14:

a: \(4n-5⋮2n-1\)

=>\(4n-2-3⋮2n-1\)

=>\(-3⋮2n-1\)

=>\(2n-1\inƯ\left(-3\right)\)

=>\(2n-1\in\left\{1;-1;3;-3\right\}\)

=>\(2n\in\left\{2;0;4;-2\right\}\)

=>\(n\in\left\{1;0;2;-1\right\}\)

mà n>=0

nên \(n\in\left\{1;0;2\right\}\)

b: \(n^2+3n+1⋮n+1\)

=>\(n^2+n+2n+2-1⋮n+1\)

=>\(n\left(n+1\right)+2\left(n+1\right)-1⋮n+1\)

=>\(-1⋮n+1\)

=>\(n+1\in\left\{1;-1\right\}\)

=>\(n\in\left\{0;-2\right\}\)

mà n là số tự nhiên

nên n=0

thiếu bài 16

\(a,\Rightarrow x=19-17=2\\ b,\Rightarrow x+8=28:2=14\\ \Rightarrow x=14-8=6\\ c,\Rightarrow42-x=5^2=25\\ \Rightarrow x=42-25=17\)

a)x=2

b)x+8=14

x=6

c)\(42-x=5^2\)

\(42-x=25\)

\(-x=-17\)

\(x=17\)

a: (x-2)(y-3)=5

=>\(\left(x-2\right)\cdot\left(y-3\right)=1\cdot5=5\cdot1=\left(-1\right)\cdot\left(-5\right)=\left(-5\right)\cdot\left(-1\right)\)

=>\(\left(x-2;y-3\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(3;8\right);\left(7;4\right);\left(1;-2\right);\left(-3;2\right)\right\}\)

b: (2x-1)*(y-4)=-11

=>\(\left(2x-1\right)\cdot\left(y-4\right)=1\cdot\left(-11\right)=\left(-11\right)\cdot1=\left(-1\right)\cdot11=11\cdot\left(-1\right)\)

=>\(\left(2x-1;y-4\right)\in\left\{\left(1;-11\right);\left(-11;1\right);\left(-1;11\right);\left(11;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(1;-7\right);\left(-5;5\right);\left(0;15\right);\left(6;3\right)\right\}\)

c: xy-2x+y=3

=>\(x\left(y-2\right)+y-2=1\)

=>\(\left(x+1\right)\left(y-2\right)=1\)

=>\(\left(x+1\right)\cdot\left(y-2\right)=1\cdot1=\left(-1\right)\cdot\left(-1\right)\)

=>\(\left(x+1;y-2\right)\in\left\{\left(1;1\right);\left(-1;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(0;3\right);\left(-2;1\right)\right\}\)