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\(\left(y+2\right)x+\left(y+2\right)=15\Leftrightarrow\left(y+2\right)\left(x+1\right)=15\)
x+1 | 1 | 3 | 5 |
y+2 | 15 | 5 | 3 |
x | 0 | 2 | 4 |
y | 13 | 3 | 1 |
3xy + x = 2y
<=> 3xy - 2y + x = 0
<=> 6(3xy - 2y + x) = 0
<=> 18xy - 12y + 6x = 0
<=> 6y(3x - 2) + 6x - 4 = - 4
<=> 6y(3y - 2) + 2(3y - 2) = - 4
<=> (6y + 2)(3y - 2) = - 4
=> (6y + 2)(3y - 2) = - 1.4 = - 2.2 = 1.(-4)
Tự liệt kê ra nhá , mỏi tay lắm
xy - 2x - 3y = 5
<=> xy - 2x - 3y + 6 = 11
<=> x(y - 2) - 3(y - 2) = 11
<=> (x - 3)(y - 2) = 11
x - 3 | 1 | -1 | 11 | -11 |
y - 2 | 11 | -11 | 1 | -1 |
x | 4 | 2 | 14 | -8 |
y | 13 | -9 | 3 | 1 |
Vậy...
a) Ta có: \(xy=-13\)
\(\Leftrightarrow x,y\inƯ\left(-13\right)\)
\(\Leftrightarrow x,y\in\left\{1;-1;13;-13\right\}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=-13\end{matrix}\right.\\\left\{{}\begin{matrix}x=-1\\y=13\end{matrix}\right.\\\left\{{}\begin{matrix}x=-13\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=13\\y=-1\end{matrix}\right.\end{matrix}\right.\)
Vậy: (x,y)∈{(1;-13);(-1;13);(-13;1);(13;-1)}
b) Ta có: \(\left(x-1\right)\left(y+2\right)=7\)
\(\Leftrightarrow x-1;y+2\inƯ\left(7\right)\)
\(\Leftrightarrow x-1;y+2\in\left\{1;7;-1;-7\right\}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x-1=1\\y+2=7\end{matrix}\right.\\\left\{{}\begin{matrix}x-1=7\\y+2=1\end{matrix}\right.\\\left\{{}\begin{matrix}x-1=-1\\y+2=-7\end{matrix}\right.\\\left\{{}\begin{matrix}x-1=-7\\y+2=-1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x=2\\y=5\end{matrix}\right.\\\left\{{}\begin{matrix}x=8\\y=-1\end{matrix}\right.\\\left\{{}\begin{matrix}x=0\\y=-9\end{matrix}\right.\\\left\{{}\begin{matrix}x=-6\\y=-3\end{matrix}\right.\end{matrix}\right.\)
Vậy: (x,y)∈{(2;5);(8;-1);(0;-9);(-6;-3)}
x.y-x-y=2
x.y-x-y+1=2+1
x.(y-1)-1.(y-1)=3
(x-1)(y-1)=3
=>x-1;y-1\(\in\)Ư(3)
Ư(3)={1,-1,3,-3}
Ta có bảng sau:
Bn tự kết luận nha!