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xy+3y-7y=21
x.(y+3)-(7y+21)
x.(y+3)-7.(y+3)=0
(y+3).(x-7)=0
suy ra: y-3=0 và y=3 và x-7=0 x=7
b, xy+3x-2y=11
x.(y+3) - (2y+11) = 0
x.(y+3) - 2.(y+5) +1 = 0
x.(y+3) - 2.(y+5) = 0-1
x.(y+3) - 2.(y+5) = -1
(y+3) . (x-7) = -1
y+3={1;-1} và x-7={1;-1}
1/ có \(xy=5\Rightarrow x,y\inƯ\left(5\right)=\left\{1,5\right\}\)
mà \(x>y\) \(\Rightarrow x=5,y=1\)
2/ \(\left(x+1\right)\left(y+2\right)=5\) \(\Rightarrow x+1,y+2\inƯ\left(5\right)=\left\{1;5\right\}\)
\(\Rightarrow\orbr{\begin{cases}x+1=5,y+2=1\Rightarrow x=4,y=-1\left(loai\right)\\x+1=1,y+2=5\Rightarrow x=0,y=3\left(tm\right)\end{cases}}\)
vậy x=0, y=3
3/ \(\left(x+1\right)\left(y+2\right)=6\) \(\Rightarrow x+1,y+2\inƯ\left(6\right)=\left\{1,2,3,6\right\}\)
=>
| x+1 | 1 | 2 | 3 | 6 |
| x | 0 | 1 | 2 | 5 |
| y+2 | 6 | 3 | 2 | 1 |
| y | 4 | 1 | 0 | -1(loại) |
vậy có 3 kết quả như bảng trên
Nhờ mọi người giải bài toán này giúp mình nha
Tìm x ,y thuộc Z và x,y >0 sao cho:
28x = 3( x^2 + 3y^2)
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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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)}
1)\(y=\frac{x^2+3x+7}{x+3}=\frac{x\left(x+3\right)+7}{x+3}=x+\frac{7}{x+3}\)= > x +3 thuoc\(U_{\left(7\right)}=\left\{1;-1;7;-7\right\}\)
x thuoc \(\left\{-2;-4;3;-11\right\}\)
2)\(y=\frac{4x+3}{2x+6}=\frac{4x+12-8}{2x+6}=\frac{2\left(2x+6\right)-8}{2x+6}=2-\frac{8}{2x+6}\) =>2x+6 thuoc
\(U_{\left(8\right)}=\left\{1;-1;2;-2;4;-4;8;-8\right\}\)
=>x thuoc \(\left\{-2;-4;-1;-5;1;-7\right\}\)
4)\(y=\frac{4x+1}{3x-1}\)
\(3y=\frac{12x+3}{3x-1}=\frac{12x-4+7}{3x-1}=\frac{4\left(3x-1\right)+7}{3x-1}=4+\frac{7}{3x-1}\)
3x+1 thuoc {1;-1;7;-7}
3x thuoc {0;-2;6;-8}
x thuoc {0;2}