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x+xy+y=9
=> x(y+1)+y+1=10
=> x(y+1)+(y+1)=10
=> (x+1)(y+1)=10
sau đó làm giống bạn Nobita Kun
ủng hộ mik nha
x + xy + y = 9
=> x(y + 1) + (y + 1) = 10
=> (x + 1)(y + 1) = 10
=> 10 chia hết cho x + 1
Ta có bảng:
x+1 | 1 | -1 | 2 | -2 | 5 | -5 | 10 | -10 |
x | 0 | -2 | 1 | -3 | 4 | -6 | 9 | -11 |
y+1 | 10 | -10 | 5 | -5 | 2 | -2 | 1 | -1 |
y | 9 | -11 | 4 | -6 | 1 | -3 | 0 | -2 |
Vậy....
![](https://rs.olm.vn/images/avt/0.png?1311)
x+xy+y=9
=>x+xy+y+1=10
=>(x+xy)+(y+1)=10
=>x.(y+1)+(y+1)=10
=>(y+1)(x+1)=10=1.10=10.1=(-1).(-10)=(-10).(-1)=2.5=5.2=(-2).(-5)=(-5).(-2)
Ta có bảng sau:
(x+1) ; (y+1) | 1;10 | 10;1 | -1;-10 | -10;-1 | 2;5 | 5;2 | -2;-5 | -5;-2 |
(x;y) | (0;9) | (9;0) | (-2;-11) | (-11;-2) | (1;4) | (4;1) | (-3;-6) | (-6;3) |
Vậy các cặp x,y tìm dc là:
(0;9) | (9;0) | (-2;-11) | (-11;-2) | (1;4) | (4;1) | (-3;-6) | (-6;3) |
Lưu ý: (a;b) thì x=a;y=b nha bạn
![](https://rs.olm.vn/images/avt/0.png?1311)
x + xy + y = 9
x(y + 1) + y = 9
x(y + 1) + y + 1 = 9 + 1
x(y+1) + (y+1) = 10
(x+1)(y+1) = 10
Bạn tự tìm
![](https://rs.olm.vn/images/avt/0.png?1311)
\(xy+x-y=4\\ \Rightarrow xy+x-y-1+1=4\\ \Rightarrow\left(xy+x\right)-\left(y+1\right)=4-1\\ \Rightarrow x\left(y+1\right)-\left(y+1\right)=3\\ \Rightarrow\left(x-1\right)\left(y+1\right)=3\\ \Rightarrow\left(x-1\right);\left(y+1\right)\inƯ\left(3\right)\\ \Rightarrow\left(x-1\right);\left(y+1\right)\in\left\{-3;-1;1;3\right\}\)
\(x-1\) | \(-3\) | \(-1\) | \(1\) | \(3\) |
\(y+1\) | \(-1\) | \(-3\) | \(3\) | \(1\) |
\(x\) | \(-2\) | \(0\) | \(2\) | \(4\) |
\(y\) | \(-2\) | \(-4\) | \(2\) | \(0\) |
Vậy \(\left(x;y\right)=\left(-2;-2\right),\left(0;-4\right),\left(2;2\right),\left(4,0\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
từ giả thiết : xy + yz = 8 ; yz + zx = 9 ; zx + xy = 5
=> xy + yz + zx = 11
=> xy = 2 ; yz = 6 ; zx = 3
=>( xyz)2 = 36 => xyz = \(\pm\)6
+ nếu xyz = 6 thì : x = 1 ; y = 2; z = 3
+ nếu xyz = -6 thì : x = -1 ; y = -2 ; z = -3
\(xy+yz=8;yz+zx=9;zx+xy=5\)
\(\Rightarrow xy+yz+yz+zx+zx+xy=8+9+5\)
\(\Leftrightarrow2xy+2yz+2xz=22\)
\(\Leftrightarrow2\left(xy+yz+xz\right)=22\)
\(\Leftrightarrow xy+yz+xz=11\)
\(\Rightarrow\hept{\begin{cases}xz=11-8\\xy=11-9\\yz=11-5\end{cases}\Rightarrow\hept{\begin{cases}xz=3\\xy=2\\yz=6\end{cases}}}\Rightarrow xz\cdot xy\cdot yz=3\cdot2\cdot6=36\)
\(\Leftrightarrow\left(xyz\right)^2=36=\left(\pm6\right)^2\)
TH1: \(xyz=6\)
\(\Rightarrow\hept{\begin{cases}xyz:xz=y\\xyz:xy=z\\xyz:yz=x\end{cases}\Rightarrow\hept{\begin{cases}y=6:3\\z=6:2\\x=6:6\end{cases}\Rightarrow}\hept{\begin{cases}y=2\\z=3\\x=1\end{cases}}}\)
TH2: \(xyz=-6\)
\(\Rightarrow\hept{\begin{cases}xyz:xz=y\\xyz:xy=z\\xyz:yz=x\end{cases}\Rightarrow\hept{\begin{cases}y=-6:3\\z=-6:2\\x=-6:6\end{cases}\Rightarrow}\hept{\begin{cases}y=-2\\z=-3\\x=-1\end{cases}}}\)
Vậy 2 tập nghiệm của x, y, z là (1;2;3) và (-1;-2;-3)
![](https://rs.olm.vn/images/avt/0.png?1311)
1.
-2...4/5....
*Khi x<-2
4-5x=-x-2=> x=6/4=3/2 loại
*khi -2<=x<4/5
4-5x=x+2=> x=2/6=1/3<4/5 nhân
*kho x>=4/5
5x-4=x+2=> x=6/4=3/2 nhận
2.
x(1+y)+y+1=10
(x+1)(y+1)=10
x+1={-10,-5,-2,-1,1,2,5,10}=> x={-11,-6,-3,-2,0,1,4,9}
y+1={-1,-2,-5,-10,10,5,2,1)=> y={-2,-3,-6,-11,9,4,1,0}
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