Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1) Ta có:
\(x+5⋮2x-1\)
\(\Rightarrow2\left(x+5\right)⋮2x-1\)
\(\Rightarrow2x+10⋮2x-1\)
\(\Rightarrow\left(2x-1\right)+11⋮2x-1\)
\(\Rightarrow11⋮2x-1\)
\(\Rightarrow2x-1\in U\left(11\right)=\left\{-1;1;-11;11\right\}\)
\(\Rightarrow\left\{{}\begin{matrix}2x-1=-1\Rightarrow x=0\\2x-1=1\Rightarrow x=1\\2x-1=-11\Rightarrow x=-5\\2x-1=11\Rightarrow x=6\end{matrix}\right.\)
Vậy \(x\in\left\{0;1;-5;6\right\}\)
1. xy + 5x + 5y = 92
=> (xy + 5x) + (5y + 25) = 92 + 25
=> x(y + 5) + 5(y + 5) = 117
=> (x + 5)(y + 5) = 117
=> x + 5 \(\in\)Ư(117) = {-1;1;-3;3;-9;9;-13;13;-39;39;-117;117}
Mà x >= 0 => x + 5 >= 5
=> x + 5 \(\in\){9;13;39;117}
Ta có bảng sau:
x + 5 | 9 | 13 | 39 | 117 |
x | 4 | 8 | 34 | 112 |
y + 5 | 13 | 9 | 3 | 1 |
y | 8 | 4 | -2 (loại) | -4 (loại) |
Vậy; (x;y) \(\in\){(4;8);(8;4)}
xy-5x-3y+8=0
x(y-5)-3(y-5)=7
(y-5)(x-3)=7
\(\Rightarrow y-5;x-3\inƯ\left(7\right)=\left\{1;-1;7;-7\right\}\)
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}y-5=1\\x-3=7\end{matrix}\right.\\\left\{{}\begin{matrix}y-5=7\\x-3=1\end{matrix}\right.\\\left\{{}\begin{matrix}y-5=-1\\x-3=-7\end{matrix}\right.\\\left\{{}\begin{matrix}y-5=-7\\x-3=-1\end{matrix}\right.\end{matrix}\right.\)
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}y=6\\x=10\end{matrix}\right.\\\left\{{}\begin{matrix}y=10\\x=6\end{matrix}\right.\\\left\{{}\begin{matrix}y=4\\x=-4\end{matrix}\right.\\\left\{{}\begin{matrix}y=-4\\x=4\end{matrix}\right.\end{matrix}\right.\)