(2x - 1)(y - 2) = -11
Ta có: -11 = (-1).11 = 1.(-11)
Ta có bảng:
| 2x - 1 | -1 | 11 | -11 | 1 |
| y - 2 | -1 | 11 | -11 | 1 |
| x | 0 | 6 | -5 | 1 |
| y | 1 | 13 | -9 | 3 |
(2x-1)(y-2)=-11
=> 2x-1; y-2 thuộc Ư (-11)={-11;-1;1;11}
Ta có bảng
| 2x-1 | -11 | -1 | 1 | 11 |
| x | -5 | 0 | 1 | 6 |
| y-2 | 1 | 11 | -11 | -1 |
| y | 3 | 13 | -9 | 1 |
mik đang cần gấp, please
(2x-1).(y-2)= -11
=> x E Ư(-11)
Ta có bảng:
| 2x-1 | 1 | -11 | -1 | 11 |
| y-2 | -11 | 1 | 11 | -1 |
| x | 1 | -5 | 0 | 6 |
| y | -9 | 3 | 13 | 1 |
Vậy.,........................................................
\(\left(2x-11\right)\left(y-2\right)=-11\)
\(Th1:2x-11=11\Leftrightarrow2x=22\Leftrightarrow x=11\)
\(y-2=-1\Leftrightarrow x=1\)
\(Th2:2x-11=1\Leftrightarrow2x=12\Leftrightarrow x=6\)
\(y-2=-11\Leftrightarrow y=-9\)
\(Th3:2x-11=-11\Leftrightarrow2x=0\Leftrightarrow x=0\)
\(y-2=1\Leftrightarrow y=3\)
\(Th4:2x-11=-1\Leftrightarrow2x=10\Leftrightarrow x=5\)
\(y-2=11\Leftrightarrow y=13\)
(2x-1).(y-2)=-11
=> (2x-1).(y-2)=1.(-11)=(-1).11=(-11).1=11.(-1)
Ta có bảng sau:
| 2x-1 | 1 | -1 | 11 | -11 |
| 2x | 2 | 0 | 12 | -10 |
| x | 1 | 0 | 6 | -5 |
| y-2 | -11 | 11 | -1 | 1 |
| y | -9 | 13 | 1 | 3 |
Vậy {x;y} \(\in\){1;-9};{0;13};{6;1};{-5;3}
\(\left(2x-1\right).\left(y-2\right)=-11\)
th1 : \(\orbr{\begin{cases}2x-1=11\\y-2=-1\end{cases}=>\orbr{\begin{cases}x=\frac{12}{2}=6\\y=-1+2=1\end{cases}}}\)
th2 : \(\orbr{\begin{cases}2x-1=-11\\y-2=1\end{cases}}=>\orbr{\begin{cases}x=\frac{-11+1}{2}=-\frac{10}{2}=-5\\y=1+2=3\end{cases}}\)
th3 : \(\orbr{\begin{cases}2x-1=1\\y-2=-11\end{cases}=>\orbr{\begin{cases}x=\frac{1+1}{2}=1\\y=-11+2=-9\end{cases}}}\)
th4 : \(\orbr{\begin{cases}2x-1=-1\\y-1=11\end{cases}=>\orbr{\begin{cases}2x=-1+1\\y=11+1\end{cases}=>\orbr{\begin{cases}x=\frac{0}{2}=0\\y=12\end{cases}}}}\)
Vậy ...
