ta có x²y -x+xy=6

=> xy(x+1)-x =6

=> xy(x+1)-1-x=6-1

=>xy(x+1)-(x+1)=5

=> (xy-1)(x+1)=5

Do \(x,y\in Z\)=> xy-1 và x+1 thuộc Ư(5)

Nên ta có bảng sau

(vì x,y là số nguyên nha bạn )

Vậy \(\left(x,y\right)\in\left\{\left(-6,0\right),\left(-2,2\right)\right\}\)

*****chúc bạn học giỏi*****

x^2y-x+xy=6

x.xy+xy-x=6

xy(x+1)-x=6

xy(x+1)-x-1=6-1

xy(x+1)-(x+1)=5

(xy-1)(x+1)=5

=> x+1=1; xy-1=5. x+1=1= > x=0 mà xy-1=5 hay 0.y-1=5 vô lí

x+1=5; xy-1=1. x+1=5 => x=4. mà xy-1=1 hay 4.y-1=1. 4y=2 thì y=1/2. y ko thuộc Z vô lí

x+1=-1; xy-1=-5. x+1=-1 => x=-2. mà xy-1=-5 hay -2.y-1=-5 => y=2.

x+1=-5; xy-1=-1. x+1=-5 =>x=-6 mà xy-1=-1 hay -6.y-1=-1 vô lí

vậy chỉ có cặp số nguyên x y thõa mãn yêu cầu: x=-2 y=2 

Ta có :x^2y-x+xy=6

x.xy +xy-x=6

xy(x+1) -x=6

xy(x+1)-(x+1)=5

(xy-1)(x+1)=5

sau đó tìm ước của 5 

a) Có lẽ đề có vấn đề.

b) \(\frac{x-11}{y-10}=\frac{11}{10}\Rightarrow10\left(x-11\right)=11\left(y-10\right)\)

\(10x-110=11y-110\)

\(10x-11y-110+110=0\)

\(10x-11y=0\)

\(10x-\left(10y+y\right)=0\)

\(10x-10y-y=0\)

\(10\left(x-y\right)-y=0\)

TH1: x-y = -12

10 (-12) -y =0

-120 - y =0

y = -120

Thay y = -120 vào x-y = -12

x - (-120) = -12

x + 120 = -12

x= -12 - 120

x= -132

TH2: x-y = 12

10 * 12 -y = 0

120 - y =0

y = 120

Thay y= 120 vào x-y = 12 

x - 120 = 12

x= 12 + 120

x= 132

Vậy nếu  y= -120 thì x= -132

nếu y= 120 thì x= 132

\(xy-x+2y=3\)

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

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

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

Ta có:1=1 . 1=(-1) . (-1)

Lập bảnh:

Vậy các cặp số ( x ; y ) là: ( -1 ; 2 ) ; ( -3 ; 0 )

\(x^2y-x+xy=6\)

\(x\left(xy-1\right)+\left(xy-1\right)=6-1\)

\(\left(x+1\right)\left(xy-1\right)=5\)

Khi \(\hept{\begin{cases}x+1=1\\xy-1=5\end{cases}\Rightarrow\hept{\begin{cases}x=0\\0-1=5\left(\text{vô lý}\right)\end{cases}}}\)

Khi \(\hept{\begin{cases}x+1=-1\\xy-1=-5\end{cases}\Rightarrow\hept{\begin{cases}x=-2\\y=2\end{cases}}}\)

Khi \(\hept{\begin{cases}x+1=5\\xy-1=1\end{cases}\Rightarrow\hept{\begin{cases}x=4\\y=\frac{1}{2}\notinℤ\end{cases}}}\)

Khi \(\hept{\begin{cases}x+1=-5\\xy-1=-1\end{cases}\Rightarrow\hept{\begin{cases}x=-6\\y=0\end{cases}}}\)

Vậy \(\left(x;y\right)\in\left\{\left(-6;0\right);\left(-2;2\right)\right\}\)

\(x^2y-x+xy=6\)

\(\Rightarrow xy\left(x+1\right)-x-1=5\)

\(\Rightarrow\left(xy-1\right)\left(x+1\right)=5\)

Lập bảng là ra

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

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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\(x=-1;y=2\)

Ta co  \(xy-x+2y=3\Rightarrow x\left(y-1\right)+2y-2=1\)

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

         \(\Rightarrow\left(y-1\right)\left(x+2\right)=3\)\(=1\times3=3\times1=\left(-1\right)\times\left(-3\right)=\left(-3\right)\times\left(-1\right)\)

Xet bang sau

Vay cac cap so \(\left(x;y\right)\)thoa man de bai la\(\left(-1;4\right),\left(1;2\right),\left(-3;-2\right),\left(-5;0\right)\)

a) \(xy+x+2y=5\Leftrightarrow xy+x+2y+2=7\Leftrightarrow\left(y+1\right)\left(x+2\right)=7\)

Vì x,y là số tự nhiên nên \(x,y\in N\)\(x,y\ge0\)\(\Rightarrow y+1\ge1;x+2\ge2\)

Từ đó ta có : 

\(\hept{\begin{cases}x+2=7\\y+1=1\end{cases}\Leftrightarrow\hept{\begin{cases}x=5\\y=0\end{cases}}}\) 

b) \(xy+2x+2y=-16\Leftrightarrow xy+2y+2x+4=-12\Leftrightarrow\left(y+2\right)\left(x+2\right)=-12\)

Lần lượt xét từng trường hợp , ta được : 

(x;y) = (-14; -1) ; (-8 ; 0) ; (-6 ; 1) ; (-5 ;2) ; (-4 ;4)

a) \(\left(x+2\right)\left(y+1\right)=7=1.7=7.1\)

Hoặc \(\hept{\begin{cases}x+2=7\\y+1=1\end{cases}\Leftrightarrow\hept{\begin{cases}x=5\\y=0\end{cases}}}\in N\)

Hoặc\(\hept{\begin{cases}x+2=1\\y+1=7\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1\notin N\\y=6\end{cases}}\)

Vậy \(\left(x;y\right)=\left(5;0\right)\)

b)\(\left(x+2\right)\left(y+2\right)=-1.12=-12.1=-2.6=-6.2=-3.4=-4.3\)

tương tự giải 6 TH là được