Trả lời nhanh và đúng thì mình k nha

xy-3x-2y=-5

=>xy-3x-2y+6=1

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

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

phần còn lại bạn tự làm nốt nha

ta có :xy-3x-2y=-5

<=>xy-2y-3x+6=1

<=>y(x-2)-3(x-2)=1

<=>(y-3)(x-2)=1 mà 1=1.1 hoặc (-1).(-1)

=>TH1: (y-3)(x-2)=1.1

=>x=3;y=4

=>TH2:(y-3)(x-2)=(-1)(-1)

=>x=1;y=2

Vậy (x;y) thuộc  {(1;2);(3;4)}

  a, xy+3x-7y=21.                                           => x(y+3) - 7(y+3) = 0.                                       => (x-7)(y+3)=0.                                               => x=7 , và với mọi y.  Hoặc y=3 với mọi x

\(a)\)

\(xy+3x-7y=21\)

\(xy+3x-7y-21=0\)

\(x\left(y+3\right)-7\left(y+3\right)=0\)

\(\Rightarrow x=7;y=\left(-3\right)\)

Vậy ...

uuttqquuậậyy gửi từng bài thì có mà hết lượt gửi câu hỏi à

a.
xy + 3x - 2y - 6 = 5
=>x(y + 3) - 2(y + 3) = 5
=>(x - 2)(y + 3) = 5.
Vì x, y thuộc Z nên x - 2, y + 3 thuộc Z
=> x - 2, y + 3 thuộc ước nguyên của 5
Lập bảng : 

Vậy ......

b. Làm tương tự câu a.

c. Ta có x + y = 3 và x - y = 15
Bài này là tổng hiệu của cấp 1, áp dụng cách làm đó thì ta được số lớn là x = (3 + 15) : 2 = 9
Số bé là y = 9 - 15 = -6

d. Ta có : |x| + |y| = 1
=>|x| = 1 - |y|
Vì |x|, |y| >= 0 và |x| = 1 - |y| nên 0 =< |x|, |y| =< 1
Vì x, y thuộc Z nên x = 0 thì y = 1 hoặc -1 và ngược lại y = 0 thì x = 1 hoặc -1

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3x + xy = 15

<=> x( 3 + y ) = 15

Vì x,y nguyên => x nguyên và 3 + y nguyên

lại có 15 = 1.15 = 3.5 = (-1).(-15) = (-3).(-5)

chổ này bạn tự làm tiếp :))

xy - 2y + 5x = 21

<=> xy - 2y + 5x - 10 = 11

<=> y( x - 2 ) + 5( x - 2 ) = 11

<=> ( x - 2 )( y + 5 ) = 11

Vì x,y nguyên => x - 2 và y + 5 nguyên

lại có 11 = 1.11 = (-1).(-11)

tương tự như a)

a) 3x + xy - y = 2

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

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

=> (x - 1)(3 + y) = 5 = 1 . 5 = 5 . 1= -1 . (-5) = (-5) . (-1)

Lập bảng : 

Vậy ...

Viết sai ùi kìa là 17-xy

3x-2y=27-xy

3x-2y+xy=27

3x+xy-2y=27

x(3+y)-2y-6=27-6

x(3+y)-(2y+6)=21

x(3+y)-2(y+3)=21

(y+3)(x-2)=21

Ta có bảng giá trị:

17 - xy nha

\(x^2+3x+5=xy+2y\\ \Leftrightarrow x^2+3x-xy-2y+5=0\\ \Leftrightarrow x\left(x+2\right)-y\left(x+2\right)+\left(x+2\right)+3=0\\ \Leftrightarrow\left(x+2\right)\left(x-y+1\right)=-3=\left(-1\right)\cdot3=\left(-3\right)\cdot1\)

\(TH_1:\left\{{}\begin{matrix}x+2=-3\\x-y+1=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-5\\y=-5\end{matrix}\right.\to\left(-5;-5\right)\\ TH_2:\left\{{}\begin{matrix}x+2=3\\x-y+1=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\to\left(1;3\right)\\ TH_3:\left\{{}\begin{matrix}x+2=1\\x-y+1=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=3\end{matrix}\right.\to\left(-1;3\right)\\ TH_4:\left\{{}\begin{matrix}x+2=-1\\x-y+1=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-5\end{matrix}\right.\to\left(-3;-5\right)\)

Vậy \(\left(x;y\right)=\left(-5;-5\right);\left(1;3\right);\left(-1;3\right);\left(-3;-5\right)\)