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\(xy=x-y\)
\(\Leftrightarrow xy-\left(x-y\right)=0\)
\(\Leftrightarrow xy-x+y=0\)
\(\Leftrightarrow x\left(y-1\right)+\left(y-1\right)=-1\)
\(\Leftrightarrow\left(y-1\right)\left(x+1\right)=-1=-1.1=1.\left(-1\right)\)
Lập bảng:
\(y-1\) | \(-1\) | \(1\) |
\(x+1\) | \(1\) | \(-1\) |
\(x\) | \(0\) | \(2\) |
\(y\) | \(0\) | \(-2\) |
Vậy \(\left(x,y\right)\in\left\{\left(0,0\right);\left(2,-2\right)\right\}\)
\(xy=x-y\)
\(\Rightarrow xy-x+y=0\)
\(\Rightarrow xy-x+y-1=-1\)
\(\Rightarrow x\left(y-1\right)+\left(y-1\right)=-1\)
\(\Rightarrow\left(x+1\right)\left(y-1\right)=-1\)
Vì \(x;y\in Z\)nên xét bảng:
x + 1 | 1 | -1 |
y - 1 | -1 | 1 |
x | 0 | -2 |
y | 0 | 2 |
Vậy \(\left(x;y\right)=\left(0;0\right)\)và \(\left(-2;2\right)\)
\(xy+x-y=4\Rightarrow x\left(y+1\right)=4+y\Leftrightarrow\frac{y+4}{y+1}=x\)
\(\Leftrightarrow y+4⋮y+1\Rightarrow3⋮y+1\)
............................
xy + x - y = 4
x.(y+1) - y - 1 = 4- 1
x.(y+1) - (y+1) = 3
(y+1).(x-1) = 3 = 3.1 = (-3).(-1)
TH1: y + 1 = 1 => y = 0 (TM)
x - 1 = 3 => x = 4 (TM)
TH2:...
TH3:...
TH4:...
bn tự xét tiếp nha!
x + xy + y = 9
<=> x + xy + y + 1 = 9 + 1
<=> x(y + 1) + (y + 1) = 10
<=> (x + 1)(y + 1) = 10
Ta có bảng sau
x + 1 | 1 | -1 | 2 | -2 | 5 | -5 | 10 | -10 |
y + 1 | -10 | 10 | -5 | 5 | -2 | 2 | -1 | 1 |
x | 0 | -2 | 1 | -3 | 4 | -6 | 9 | -11 |
y | -11 | 9 | -6 | 4 | -3 | 1 | -2 | 0 |
Vậy các cặp (x;y) thõa mãn là (0;-11) ; (-2;9) ; (1;-6) ; (-3;4) ; (4;-3) ; (-6;1) ; (9;-2) ; (-11;0)
May ngu
Tao lv 1211 lc 100k ma moi v111
TaoTM
may la hinata
T
XIn loi ban minh len con dong kinh
\(xy+12=x-y\)
\(\Rightarrow xy-x+y=-12\)
\(\Rightarrow x\left(y-1\right)+\left(y-1\right)=-13\)
\(\Rightarrow\left(y-1\right)\left(x+1\right)=-13\)
\(\Rightarrow\left(y-1\right)\left(x+1\right)=-1.13=-13.1=1.\left(-13\right)=13.\left(-1\right)\)
Đến đây bn lập bảng lak tìm ra x,y
\(xy+12=x-y\)
\(\Rightarrow xy-x+y=-12\)
\(\Rightarrow x\left(y-1\right)+\left(y-1\right)=-12-1\)
\(\Rightarrow\left(x+1\right)\left(y-1\right)=-13\)
\(\Rightarrow\left(x+1\right)\left(1-y\right)=13\)
\(\Rightarrow\left(x+1\right);\left(1-y\right)\inƯ\left(13\right)=\left\{\pm1;\pm13\right\}\)
Xét bảng
x+1 | 1 | -1 | 13 | -13 |
1-y | 13 | -13 | 1 | -1 |
x | 0 | -2 | 12 | -14 |
y | -12 | 14 | 0 | 2 |
Vậy......................................
a: xy=x-y
=>xy-x+y=0
=>xy-x+y-1=-1
=>x(y-1)+(y-1)=-1
=>(x+1)(y-1)=-1
=>\(\left(x+1\right)\left(y-1\right)=1\cdot\left(-1\right)=\left(-1\right)\cdot1\)
=>\(\left(x+1;y-1\right)\in\left\{\left(1;-1\right);\left(-1;1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(0;0\right);\left(-2;2\right)\right\}\)
b: x(y+2)+y=1
=>\(x\left(y+2\right)+y+2=3\)
=>\(\left(x+1\right)\left(y+2\right)=3\)
=>\(\left(x+1\right)\cdot\left(y+2\right)=1\cdot3=3\cdot1=\left(-1\right)\left(-3\right)=\left(-3\right)\left(-1\right)\)
=>\(\left(x+1;y+2\right)\in\left\{\left(1;3\right);\left(3;1\right);\left(-1;-3\right);\left(-3;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(0;1\right);\left(2;-1\right);\left(-2;-5\right);\left(-4;-3\right)\right\}\)
a) \(xy+x+y=2\)
\(xy+x+y+1=2+1\)
\(\left(xy+x\right)+\left(y+1\right)=3\)
\(x\left(y+1\right)+\left(y+1\right)=3\)
\(\left(y+1\right)\left(x+1\right)=3\)
\(\Rightarrow\left\{{}\begin{matrix}x+1\in\left\{-3;-1;1;3\right\}\\y+1\in\left\{-1;-3;3;1\right\}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x\in\left\{-4;-2;0;2\right\}\\y\in\left\{-2;-4;2;0\right\}\end{matrix}\right.\)
Vậy ta tìm được các cặp giá trị \(\left(x;y\right)\) thỏa mãn yêu cầu:
\(\left(-4;-2\right);\left(-2;-4\right);\left(0;2\right);\left(2;0\right)\)
b) \(\left(x+1\right).y+2=-5\)
\(\left(x+1\right).y=-5-2\)
\(\left(x+1\right).y=-7\)
\(\Rightarrow\left\{{}\begin{matrix}x+1\in\left\{-7;-1;1;7\right\}\\y\in\left\{1;7;-7;-1\right\}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x\in\left\{-8;-2;0;6\right\}\\y\in\left\{1;7;-7;-1\right\}\end{matrix}\right.\)
Mà \(x< y\)
\(\Rightarrow\left\{{}\begin{matrix}x\in\left\{-8;-2\right\}\\y\in\left\{1;7\right\}\end{matrix}\right.\)
Vậy ta tìm được các cặp giá trị \(\left(x;y\right)\) thỏa mãn yêu cầu:
\(\left(-8;1\right);\left(-2;7\right)\)
\(xy+y+x=-3\)
\(\Leftrightarrow y\left(x+1\right)+x=-3\)
\(\Leftrightarrow y\left(x+1\right)+x+1=-3+1\)
\(\Leftrightarrow\left(x+1\right)\left(y+1\right)=-2\)
Lập bảng giá trị ta có:
Vậy các cặp giá trị \(\left(x;y\right)\)thoả mãn là \(\left(-2;1\right)\), \(\left(-3;0\right)\), \(\left(0;-3\right)\), \(\left(1;-2\right)\)
xy + y + x = -3
=> xy + y +x = -1 - 2
=> xy + y + x + 1 = -2
=> y (x + 1) + (x + 1) = -2
=> (x + 1) (y + 1) = -2
Mà -2 = (-1) .2 = (-2) . 1 = 2 . (-1) = ( x + 1) (y+ 1) = (-1) . 2 = 2. (-1) = 1. (-2) = (-2) . 1 = 1. (-2)
Ta có bảng:
Vậy (x,y) = {(-3,0); (-2;4) (0 ; 3) ; (1 ; -2) }