Lời giải:

a. $2y(3x-1)+9x-3=7$

$2y(3x-1)+3(3x-1)=7$

$(3x-1)(2y+3)=7$

Vì $3x-1, 2y+3$ đều là số nguyên với mọi $x,y\in N$, và $2y+3>0$ nên ta có bảng sau:

b.

$3xy-2x+3y-9=0$

$x(3y-2)+3y-9=0$

$x(3y-2)+(3y-2)-7=0$

$(3y-2)(x+1)=7$

Đến đây bạn cũng lập bảng tương tự như phần a.

\(a,=2x^2-\dfrac{3}{2}y+3x\)

\(b,\)bt để chia hết cho x+2 là:\(2x^3+x^2-x+10\)

\(\Rightarrow m=12\)

3xy - 5y + 6x = 30

<=> y(3x - 5) + (6x - 10) = 20

<=> y(3x - 5) + 2(3x - 5) = 20

<=> (3x - 5)(y + 2) = 20

Ta có bảng sau:

Study well

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Xin lỗi phải là các cặp số nguyên x,y

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 \(x\in\left\{-1;-2;5;-8\right\}\)

a: =(a^2-b^2)-(2a-2b)

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

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

b: =(3x-3y)+5y(x-y)

=3(x-y)+5y(x-y)

=(x-y)(5y+3)

c: \(=\left(x+y\right)^2\left(x-y\right)+x\left(y-x\right)\)

=(x-y)*(x+y)^2-x(x-y)

=(x-y)[(x+y)^2-x]

d: \(=\left(x-y+4-2x-3y+1\right)\left(x-y+4+2x+3y-1\right)\)

=(-x-4y+5)(3x+2y+3)

e: =16-(x^2-4xy+4y^2)

=16-(x-2y)^2

=(4-x+2y)(4+x-2y)

g: =9x^2-6x+1-(3xy-y)

=(3x-1)^2-y(3x-1)

=(3x-1)(3x-y-1)

h: =(x-y)^3-z^3

=(x-y-z)[(x-y)^2+z(x-y)+z^2]

=(x-y-z)(x^2-2xy+y^2+xz-yz+z^2)

a) \(a^2-b^2-2a+2b\)

\(=\left(a^2-b^2\right)-\left(2a-2b\right)\)

\(=\left(a+b\right)\left(a-b\right)-2\left(a-b\right)\)

\(=\left(a-b\right)\left(a+b-2\right)\)

b) \(3x-3y-5x\left(y-x\right)\)

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

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

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

c) \(x\left(x+y\right)^2-y\left(x+y\right)^2+xy-x^2\)

\(=\left(x+y\right)^2\left(x-y\right)+\left(xy-x^2\right)\)

\(=\left(x+y\right)^2\left(x-y\right)-x\left(x-y\right)\)

\(=\left(x-y\right)\left(x^2+2xy+y^2-x\right)\)

d) \(\left(x-y+4\right)^2-\left(2x+3y-1\right)\)

\(=\left(x-y+4+2x+3y-1\right)\left(x-y+4-2x-3y+1\right)\)

\(=\left(3x+2y+3\right)\left(-x-4y+5\right)\)