Có 2 TH

\(TH1:3x>y\)

\(\Rightarrow xy+3x-y=6\)

\(\Rightarrow x\left(y+3\right)-y-3=6-3=3\)

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

Ta có bảng sau :

Vậy có các cặp (x;y)=(2;0);(4;-2);(0;-6);(-2;-4)

\(TH2:3x< y\)

\(\Rightarrow xy+y-3x=6\)

\(\Rightarrow x\left(y-3\right)+y=6\)

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

Ta có bảng sau :

Vậy ta có các cặp (x;y)=(0;6);(2;4);(-2;0);(-4;2)

\(TH1:x\ge\frac{y}{3}\) PT có dạng : \(xy+3x-y=6\)

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

Lập bảng hoặc xét từng giá trị ta được \(\left(x;y\right)=\left\{\left(2;0\right);\left(0;-6\right);\left(4;-2\right)\right\}\)

\(TH2:x< \frac{y}{3}\) Tương tự

Trường hợp 1: \(3x>y\)

\(xy+\left|3x-y\right|=6\)

\(\Leftrightarrow xy+3x-y=6\)

\(\Leftrightarrow x\left(y+3\right)-y-3=6-3\)

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

\(\Leftrightarrow\left(y+3\right).\left(x-1\right)=3\)

\(\Rightarrow y+3\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)

Mà \(y+3\ge0+3=3\)với mọi \(y\in N\)

\(\Rightarrow y+3=3\Leftrightarrow y=0\)

\(\Rightarrow x-1=3:3=1\Leftrightarrow x=2\)

Khi đó, \(3.x=6>y=0\)( thỏa mãn )

Trường hợp 2: \(3x< y\)

\(xy+\left|3x-y\right|=6\)

\(\Leftrightarrow xy+\left[-\left(3x-y\right)\right]=6\)

\(\Leftrightarrow xy-3x+y=6\)

\(\Leftrightarrow x.\left(y-3\right)+y-3=6-3\)

\(\Leftrightarrow\left(y-3\right).\left(x+1\right)=3\)

\(\Rightarrow x+1\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)

Do \(x+1\ge0+1=1\left(x\in N\right)\Rightarrow x+1\in\left\{1;3\right\}\)

\(\Rightarrow x\in\left\{0;2\right\}\)

\(\Rightarrow\orbr{\begin{cases}y-3=3:1=3\\y-3=3:3=1\end{cases}\Leftrightarrow\orbr{\begin{cases}y=6\\y=4\end{cases}}}\)

Để \(3x< y\Rightarrow\orbr{\begin{cases}x=0;y=4\\x=0;y=6\end{cases}}\)

Nhưng thử lại ta thấy có x = 0; y = 6 mới thỏa mãn.

Qua hai trường hợp, ta có:

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

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

x+y+xy=11

=> x(y+1)+y=11

=> x(y+1)+y+1=12

=> (y+1)(x+1)=12=1.12;2.6;3.4;4.3;6.2;12.1

Vậy: (x;y)= (11;0);(0;11);(5;1);(1;5);(2;3);(3;2)


 

Còn x+6=y(x-1)

Hưỡng dẫn: Đổi vế x từ x+6 sau đó trừ cả vế với 1 nhóm lại ta được hai thừa số

Sau đó lập thành tích.

Cậu biết giải rồi mà còn đăng bài làm gì nữa? Để khoe hay sao vậy?

a)Do x,y là STN mà xy=6=1.6=2.3

=>(x;y)={(1;6);(6;1);(2;3);(3;2)}

b)Do x,y là STN mà xy=40=1.40=2.20=4.10=8.5

=>(x;y)={(1;40);(40;1);(2;20);(20;2);(4;10);(10;4);(8;5);(5;8)}

c) \(^{x^2}\)+xy-x-2=0

bn vào trang wed này mik chỉ cho, cứ nhắn tin cho mik đi rồi mik sẽ ns.

x(y+2)+3y =6

=>x(y+3)+3y+9=15

=>x(y+3)+3(y+3)=15

=>(x+3)(y+3)=15

mả .....=......=>ta co bang sau

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