b.(a+b)-(b-a)+c=2a+c

Xét VT: (a+b)-(b-a)+c = a + b - b + a + c = 2a+c

Mà VP = 2a+c

=> VT = VP 

c.-(a+b-c)+(a-b-c)=-2b

Xét VT: -(a+b-c)+(a-b-c) = -a - b + c + a - b - c = -2b

Mà VP = -2b

=> VT = VP

d.a(b+c)-a(b+d)=a(c-d)

Xét VT: a(b+c)-a(b+d) = ab + ac - ab - ad =  ac - ad = a(c-d)

Mà VP = a(c-d)

=> VT = VP

e.a(b-c)+a(d+c)=a(b+d)

Xét VT: a(b-c)+a(d+c)= ab -ac + ad + ac = ab + ad = a(b+d)

Mà VP = a(b+d)

=> VT = VP

a) ta có 

abcd=120 mà abc=-30 nên -30.d=120 suy ra d=-4

abc=-30 mà ab=-6 nên -6.c=-30 suy ra c=5

bc=-15 mà c=5 suy ra b=-3

ab=-6 mà b=-3 suy ra a.(-3) = -6 suy ra a=2

b) a+b=-1, a+c=6, b+c=1 nên 2a + 2b+2c= -1 + 6 + 1 = 6

suy ra a+b+c = 3 mà a+b= -1 suy ra c=4

suy ra a=6-4=2; b=1-4 = -3

c) a+b+c=-6, b+c+d = -9, c+d+a = -8, d+a+b = -7 nên 3a+3b+3c+3d = -30

suy ra a+b+c+d= -10

mà a+b+c = -6 

suy ra d=-4

nên b+c=5, a+c=-4, a+b = -3 suy ra 2a+2b+2c = -2 suy ra a+b+c=-1

suy ra a=-6, b= 3, c= 2

a, d=-4     c=5     b=-3     a=2

b, c=4     a=2      b=-3

c, d=-4   a=-1     c=-3    b=-2

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

Bài 16:

a: \(\left(x+5\right)\left(y-3\right)=15\)

=>\(\left(x+5\right)\left(y-3\right)=1\cdot15=15\cdot1=\left(-1\right)\cdot\left(-15\right)=\left(-15\right)\cdot\left(-1\right)=3\cdot5=5\cdot3=\left(-3\right)\cdot\left(-5\right)=\left(-5\right)\cdot\left(-3\right)\)

=>\(\left(x+5;y-3\right)\in\left\{\left(1;15\right);\left(15;1\right);\left(-1;-15\right);\left(-15;-1\right);\left(3;5\right);\left(5;3\right);\left(-3;-5\right);\left(-5;-3\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(-4;18\right);\left(10;4\right);\left(-6;-12\right);\left(-20;2\right);\left(-2;8\right);\left(0;6\right);\left(-8;-2\right);\left(-10;0\right)\right\}\)

mà (x,y) là cặp số tự nhiên

nên \(\left(x,y\right)\in\left\{\left(10;4\right);\left(0;6\right)\right\}\)

b: x là số tự nhiên

=>2x-1 lẻ và 2x-1>=-1

\(\left(2x-1\right)\left(y+2\right)=24\)

mà 2x-1>=-1 và 2x-1 lẻ

nên \(\left(2x-1\right)\cdot\left(y+2\right)=\left(-1\right)\cdot\left(-24\right)=1\cdot24=3\cdot8\)

=>\(\left(2x-1;y+2\right)\in\left\{\left(-1;-24\right);\left(1;24\right);\left(3;8\right)\right\}\)

=>\(\left(2x;y\right)\in\left\{\left(0;-26\right);\left(2;22\right);\left(4;6\right)\right\}\)

=>\(\left(x;y\right)\in\left\{\left(0;-26\right);\left(1;11\right);\left(2;6\right)\right\}\)

mà (x,y) là cặp số tự nhiên

nên \(\left(x,y\right)\in\left\{\left(1;11\right);\left(2;6\right)\right\}\)

c:

x,y là các số tự nhiên

=>x+3>=3 và y+2>=2

xy+2x+3y=0

=>\(xy+2x+3y+6=6\)

=>\(x\left(y+2\right)+3\left(y+2\right)=6\)

=>\(\left(x+3\right)\left(y+2\right)=6\)

mà x+3>=3 và y+2>=2

nên \(\left(x+3\right)\cdot\left(y+2\right)=3\cdot2\)

=>x=0 và y=0

d: xy+x+y=30

=>\(xy+x+y+1=31\)

=>\(x\left(y+1\right)+\left(y+1\right)=31\)

=>\(\left(x+1\right)\left(y+1\right)=31\)

\(\Leftrightarrow\left(x+1\right)\cdot\left(y+1\right)=1\cdot31=31\cdot1=\left(-1\right)\cdot\left(-31\right)=\left(-31\right)\cdot\left(-1\right)\)

=>\(\left(x+1;y+1\right)\in\left\{\left(1;31\right);\left(31;1\right);\left(-1;-31\right);\left(-31;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(0;30\right);\left(30;0\right);\left(-2;-32\right);\left(-32;-2\right)\right\}\)

mà (x,y) là cặp số tự nhiên

nên \(\left(x,y\right)\in\left\{\left(0;30\right);\left(30;0\right)\right\}\)

b: \(\Leftrightarrow x+8\in\left\{1;-1;5;-5\right\}\)

hay \(x\in\left\{-7;-9;-3;-13\right\}\)

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

e) 

\(x-\left(17-8\right)=5+\left(10-3x\right)\\ \Rightarrow x-9=5+10-3x\\ \Rightarrow x+3x=5+10+9\\ \Rightarrow4x=24\\ \Rightarrow x=\dfrac{24}{4}=6\)

Vậy \(x=6\)