\(a,\left(x+3\right)\left(y+2\right)=1\)
=> x+3 và y+2 thuộc UC(1)={1; -1}
| x+3 | 1 | -1 |
| x | -2 | -4 |
| y+2 | 1 | -1 |
| y | -1 | -3 |
Vậy x=-2; y=-4
x=-1; y=-4
Câu sau tương tự
\(a,\left(x+3\right)\left(y+2\right)=1\)
Th1 : \(\hept{\begin{cases}x+3=1\\y+2=1\end{cases}\Rightarrow\hept{\begin{cases}x=-2\\y=-1\end{cases}}}\)
Th2 : \(\hept{\begin{cases}x+3=-1\\y+2=-1\end{cases}\Rightarrow\hept{\begin{cases}x=-4\\y=-3\end{cases}}}\)
KL : \(\left\{\left(x=-2;y=-1\right);\left(x=-4;y=-3\right)\right\}\)
\(d,3x+4y-xy=16\)
\(=3x-xy+4y-12=4\)
\(\Rightarrow-x\left(y-3\right)+4\left(y-3\right)=4\)
\(\Rightarrow\left(y-3\right)\left(4-x\right)=4\)
Chia các trường hợp như câu a của chị ra em nhé
vì \(\left(x+3\right)\left(y+2\right)=1\)
\(\Rightarrow\left(x+3\right)\text{và}\left(y+2\right)=1\)
\(x+3=1\) \(y+2=1\)
\(x=1-3\) \(y=1-2\)
\(x=-2\) \(y=-1\)
à bn ơi câu a mik thiếu
câu b mik làm nhaa
\(\Rightarrow\left(x-1\right)\text{và}\left(x+y\right)\inƯ\left(33\right)\)
\(Ư\left(33\right)=\left\{\pm1;\pm3;\pm11\pm33\right\}\)
TH1: TH2:
\(x-1=1\) \(x-1=-1\)
\(x=2\) \(x=-1+1\)
\(x+y=33\) \(x=0\)
thay số \(x=2\) \(x+y=-33\)
\(2+y=33\) thay số:\(x=0\)
\(y=33-2\) \(0+y=-33\)
\(y=31\) \(y=-33-0\)
\(y=-33\)
TH3: TH4:
\(x-1=33\) \(x-1=-33\)
\(x=33+1\) \(x=-33+1\)
\(x=34\) \(x=-32\)
\(x+y=1\) \(x+y=-1\)
thay số \(x=34\) thay số:\(x=-32\)
\(34+y=1\) \(-32+y=-1\)
\(y=1-34\) \(y=-1+32\)
\(y=-33\) \(y=31\)
TH5: TH6:
\(x-1=3\) \(x-1=-3\)
\(x=3+1\) \(x=-3+1\)
\(x=4\) \(x=-2\)
\(x+y=11\) \(x+y=-11\)
thay số : \(x=4\) thay số:\(x=-2\)
\(4+y=11\) \(-2+y=-11\)
\(y=11-4\) \(y=-11+2\)
\(y=7\) \(y=-9\)
TH7: TH8:
\(x-1=-11\) \(x-1=11\)
\(x=-11+1\) \(x=11+1\)
\(x=-10\) \(x=12\)
\(x+y=-11\) \(x+y=3\)
thay số : \(x=-10\) \(12+y=3\)
\(-10+y=-3\) \(y=3-12\)
\(y=-3+10\) \(y=-11\)
\(y=7\)
