3xy-5y+6x=30

3x(y+2)-5y=30

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

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

\(\orbr{\begin{cases}3x-5=\left\{...-2,-1,1,2,4,5,10,20\right\}\\y+2=\left\{-10,-20,20,10,5,4,2,1\right\}\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{5+\left\{...\right\}}{3}=\left\{2,3,5\right\}\\y=\left\{..\right\}-2=\left\{18,3,0\right\}\end{cases}}\)

đây là tìm số tự nhiên mà bn

=>9xy-6x+15y=87

=>3x(3y-2)+15y=87

=>3x(3y-2)+15y-10=77

=>3x(3y-2)+5(3y-2)=77

(3y-2)(3x+5)=77

đến đây bạn tự giải nhé

dễ lắm, chỉ cần phân tích 30= 6.5, 5.6, ,3.10,10.3,... rồi giải thôi

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:

A, Ta có : 2xy + x + y = 7

=> 2(2xy + x + y) = 2 . 7

=> 4xy + 2x + 2y = 14

=> (4xy + 2x) + 2y + 1 = 14 + 1

=> 2x(2y + 1) + (2y + 1) = 15

=> (2x + 1)(2y + 1) = 15

=> 2x + 1;2y + 1 ∈ Ư(15) ∈ {-15;-5;-3;-1;1;3;5;15}

Vậy ta có bảng : 

=> (x;y) = (-8;-1);(-1;-8);(-2;-3);(-3;-2);(7;0);(0;7);(1;2);(2;1)

MK HOK LOP SAU DA PHAI LAM RUI HUHU

a, 3x ( y+1) + y + 1 = 7

(y+1)(3x +1) =7

th1 : \(\left\{{}\begin{matrix}y+1=1\\3x+1=7\end{matrix}\right.\) => \(\left\{{}\begin{matrix}y=0\\x=2\end{matrix}\right.\)

th2: \(\left\{{}\begin{matrix}y+1=-1\\3x+1=-7\end{matrix}\right.\)=> x = -8/3 (loại)

th3: \(\left\{{}\begin{matrix}y+1=7\\3x+1=1\end{matrix}\right.\)=> \(\left\{{}\begin{matrix}y=6\\x=0\end{matrix}\right.\)

th 4 : \(\left\{{}\begin{matrix}y+1=-7\\3x+1=-1\end{matrix}\right.\)=> x=-2/3 (loại)

Vậy (x,y)= (2 ;0);  (0; 6)

b, xy - x + 3y - 3 = 5

   (x( y-1) + 3( y-1) = 5

          (y-1)(x+3) = 5

 th1: \(\left\{{}\begin{matrix}y-1=1\\x+3=5\end{matrix}\right.\) => \(\left\{{}\begin{matrix}y=2\\x=8\end{matrix}\right.\)

th2: \(\left\{{}\begin{matrix}y-1=-1\\x+3=-5\end{matrix}\right.\) => \(\left\{{}\begin{matrix}y=0\\x=-8\end{matrix}\right.\)

th3: \(\left\{{}\begin{matrix}y-1=5\\x+3=1\end{matrix}\right.\) => \(\left\{{}\begin{matrix}y=6\\x=-2\end{matrix}\right.\)

th4: \(\left\{{}\begin{matrix}y-1=-5\\x+3=-1\end{matrix}\right.\) =>  \(\left\{{}\begin{matrix}y=-4\\x=-4\end{matrix}\right.\)

vậy (x, y) = ( 8; 2); ( -8; 0);  (-2; 6); (-4; -4)

c, 2xy + x + y = 7 => y = \(\dfrac{7-x}{2x+1}\) ; y ϵ Z ⇔ 7-x ⋮ 2x+1

⇔ 14 - 2x ⋮ 2x + 1 ⇔ 15 - 2x - 1  ⋮ 2x + 1

th1 : 2x + 1 = -1=> x = -1; y = \(\dfrac{7-(-1)}{-1.2+1}\) = -8

th2: 2x+ 1 = 1=> x =0; y = 7

th3: 2x+1 = -3 => x =  x=-2  => y = \(\dfrac{7-(-2)}{-2.2+1}\) = -3 

th4: 2x+ 1 = 3 => x = 1 => y = \(\dfrac{7+1}{2.1+1}\) = 2

th5: 2x + 1 = -5 => x = -3=> y = \(\dfrac{7-(-3)}{-3.2+1}\) = -2

th6: 2x + 1 = 5 => x = 2; ; y = \(\dfrac{7-2}{2.2+1}\) =1

th7 : 2x + 1 = -15 => x = -8; y = \(\dfrac{7-(-8)}{-8.2+1}\) = -1

th8 : 2x+1 = 15 => x = 7; y = \(\dfrac{7-7}{2.7+1}\) = 0

kết luận

(x,y) = (-1; -8); (0 ;7); ( -2; -3) ; ( 1; 2); ( -3; -2); (2;1); (-8;-1);(7;0)

3xy−2x+5y=293xy−2x+5y=29

9xy−6x+15y=879xy−6x+15y=87

(9xy−6x)+(15y−10)=77(9xy−6x)+(15y−10)=77

3x(3y−2)+5(3y−2)=773x(3y−2)+5(3y−2)=77

(3y−2)(3x+5)=77(3y−2)(3x+5)=77

⇒(3y−2)⇒(3y−2) và (3x+5)(3x+5) là Ư(77)=±1,±7,±11,±77Ư(77)=±1,±7,±11,±77

Ta có bảng giá trị sau:

Do x,y∈Zx,y∈Z nên (x,y)∈{(−4;−3),(−2;−25),(2;3),(24;1)}

Lời giải:

$3xy-2x+5y=9$

$x(3y-2)+5y=9$

$3x(3y-2)+15y=27$

$3x(3y-2)+5(3y-2)=17$
$(3x+5)(3y-2)=17$

Do $x,y$ nguyên nên $3x+5, 3y-2$ cũng là số nguyên. Ta có bảng sau:

(2x + 1) . (2x + 2) . (2x + 3) . (2x + 4) - 5y = 11879

[(2x + 1). (2x + 4)].[(2x + 2) . (2x + 3)] -5y = 11879

(4x²+10x+4).(4x²+10x+6) -5y = 11879

Đặt t= 4x²+10x+4

t(t+2) -5y = 11879

t²+2t-5y = 11879

(t+1)² = 11880+5y

(4x²+10x+5)² = 5(2376+y)

=> x = 0; y=-2371

thanks

giải chi tiết cho mình từ chỗ thứ 8 đi

x+5=15 vây x=10

y-3=15 vậy y=18