xy - 2x = 5
x . ( y - 2 ) = 5
=> x và ( y - 2 ) ∈ Ư(5)
Mà Ư(5) = { 1 ; 5 ; - 1 ; - 5 }
TH1: x = 1 ; ( y - 2 ) = 5
y - 2 = 5
y = 5 + 2
y = 7
TH2: x = 5 ; ( y - 2 ) = 1
y - 2 = 1
y = 1 + 2
y = 3
TH3: x = - 1 ; ( y - 2 ) = - 5
y - 2 = - 5
y = - 5 + 2
y = - 3
TH4: x = - 5 ; ( y - 2 ) = - 1
y - 2 = - 1
y = - 1 + 2
y = 1
Vậy x ∈ { 1 ; 5 ; - 1 ; - 5 }
y ∈ { 7 ; 3 ; - 3 ; 1 }
\(\text{xy - 2x = 5}\)
\(\text{x . ( y - 2 ) = 5}\)
\(\Rightarrow x.\left(y-2\right)=5.1=1.5=\left(-1\right).\left(-5\right)=\left(-5\right).\left(-1\right)\)
| x | 5 | 1 | -1 | -5 |
| y - 2 | 1 | 5 | -5 | -1 |
| y | 3 | 7 | -3 | 1 |
\(\left(x,y\right)\in\left\{\left(5;3\right),\left(1;7\right),\left(-1;-3\right),\left(-5;1\right)\right\}\)
\(xy-2x=5\Leftrightarrow x\left(y-2\right)=5\)
Do vế phải dương nên x và y - 2 đồng dấu
Ta có bảng:
| x | 5 | 1 | -1 | -5 |
| y -2 | 1 | 5 | -5 | -1 |
| y | 3 | 7 | -3 | 1 |
Vậy (x;y) = (5;3) ; (1;7) ; (-1;-3) ; (-5;1)
\(xy-2x=5\)
\(\Rightarrow x\left(y-2\right)=5\)
\(\Rightarrow x,\left(y-2\right)\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
\(\Rightarrow TH1:\hept{\begin{cases}x=1\\y-2=5\end{cases}\Rightarrow\hept{\begin{cases}x=1\\y=7\end{cases}}}\) \(TH2:\hept{\begin{cases}x=-1\\y-2=-5\end{cases}\Rightarrow\hept{\begin{cases}x=-1\\y=-3\end{cases}}}\)
\(TH3:\hept{\begin{cases}x=5\\y-2=1\end{cases}\Rightarrow\hept{\begin{cases}x=5\\y=3\end{cases}}}\) \(TH4:\hept{\begin{cases}x=-5\\y-2=-1\end{cases}\Rightarrow\hept{\begin{cases}x=-5\\y=-1\end{cases}}}\)
Vậy ..............................
\(xy-2x=5\)
\(\Leftrightarrow x\left(y-2\right)=5\)
\(\Leftrightarrow x\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\)
\(\text{Trường hợp : }x=-5\)
\(\Leftrightarrow-5\left(v-2\right)=5\)
\(\Leftrightarrow y-2=5\div\left(-5\right)\)
\(\Leftrightarrow y-2=-1\)
\(\Leftrightarrow y=-1+2\)
\(\Leftrightarrow y=1\)
\(\text{Trường hợp : }x=-1\)
\(\Leftrightarrow-1\left(y-2\right)=5\)
\(\Leftrightarrow y-2=5\div\left(-1\right)\)
\(\Leftrightarrow y-2=-5\)
\(\Leftrightarrow y=-5+2\)
\(\Leftrightarrow y=-3\)
\(\text{Trường hợp : }x=1\)
\(\Leftrightarrow1\left(y-2\right)=5\)
\(\Leftrightarrow y-2=5\)
\(\Leftrightarrow y=5+2\)
\(\Leftrightarrow y=7\)
\(\text{Trường hợp : }x=5\)
\(\Leftrightarrow5\left(y-2\right)=5\)
\(\Leftrightarrow y-2=1\)
\(\Leftrightarrow y=1+2\)
\(\Leftrightarrow y=3\)
\(\text{Vậy }x,y\in\left\{\left(5,1\right);\left(-1,-3\right);\left(1,7\right);\left(5,3\right)\right\}\)
