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21 tháng 1 2017
a) \(xy+2x+3y=-6\)
\(\Rightarrow x\left(y+2\right)+3y+6=0\)
\(\Rightarrow x\left(y+2\right)+3\left(y+2\right)=0\)
\(\Rightarrow\left(x+3\right)\left(y+2\right)=0\)
\(\Rightarrow\left[\begin{matrix}x+3=0\\y+2=0\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=-3\\y=-2\end{matrix}\right.\)
Vậy \(x=-3;y=-2\)
21 tháng 1 2017
\(xy+2x+3y=-6\)
\(\Leftrightarrow xy+2x+3y+6=0\)
\(\Leftrightarrow y\left(x+3\right)+\text{2}\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(y+2\right)=0\)
\(\Leftrightarrow\left\{\begin{matrix}x+3=0\\y+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{\begin{matrix}x=-3\\y=-2\end{matrix}\right.\)
Vậy \(\left\{\begin{matrix}x=-3\\y=-2\end{matrix}\right.\)
NH
2
S
24 tháng 3 2017
xy - 5x - 3y = -8
<=> x(y - 5) - 3y + 15 = -8 + 15
<=> x(y - 5) - 3(y - 5) = 7
<=> (x - 3)(y - 5) = 7
=> x - 3 và y - 5 thuộc Ư(7) = {1;-1;7;-7}
Ta có bảng:
| x - 3 | 1 | -1 | 7 | -7 |
| y - 5 | 7 | -7 | 1 | -1 |
| x | 4 | 2 | 10 | -4 |
| y | 12 | -2 | 6 | 4 |
Vậy các cặp (x;y) là (4;12) ; (2;-2) ; (10;6) ; (-4;4)
24 tháng 3 2017
xy - 5x - 3y = -8
x (y - 5) - 3y + 15 = 7
x (y - 5 ) - 3 (y - 5) = 7
(y - 5)(x - 3) = 7
Lập bảng:
| y - 5 | 1 | -1 | 7 | -7 |
| x - 3 | 7 | -7 | 1 | -1 |
| y | 6 | 4 | 12 | -2 |
| x | 10 | -4 | 4 | 2 |
DQ
2
18 tháng 3 2017
a) xy - x - y = 10 => (xy - x) - (y - 1) = 11 => (x - 1)(y - 1) = 11 => Tự bạn giải tiếp nha
b) xy + 3x - 6y = 21 => (xy + 3x) - (6y + 18) = 3 => (x - 6)(y + 3) = 3 => Tự bạn giải tiếp nha
c) xy + 4x - 3y =12 => (xy + 4x) - (3y + 12) = 0 => (x - 3)(y + 4) = 0 => x = 3 hoặc y = -4
18 tháng 3 2017
a,xy-x-y=10
=>x(y-1)-y+1=10+1
=>x(y-1)-1(y-1)=11
=>(x-1)(y-1)=11
=>x-1 va y-1 la uoc cua 11
................
hai y con lai lam giong nhu vay
26 tháng 1 2018
a, Vì |2x+8| và |3y-9x| đều >= 0
=> |2x+8| + |3y-9x| >= 0
Dấu "=" xảy ra <=> 2x+8=0 và 3y-9x=0 <=> x=-4 và y=-12
Vậy x=-4 và y=-12
Tk mk nha
PN
0
N
9 tháng 12 2016
x(y+2)+3y =6
=>x(y+3)+3y+9=15
=>x(y+3)+3(y+3)=15
=>(x+3)(y+3)=15
mả .....=......=>ta co bang sau
4 tháng 11 2024
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PT
5
21 tháng 2 2019
\(xy+x-y=6\)
\(\Rightarrow x\left(y+1\right)-\left(y+1\right)=5\)
\(\Rightarrow\left(x-1\right)\left(y+1\right)=5\)
Làm nốt
\(xy+2x+3y=-6\)
\(\Rightarrow x\left(y+2\right)+3y+6=0\)
\(\Rightarrow x\left(y+2\right)+3\left(y+2\right)=0\)
\(\Rightarrow\left(x+3\right)\left(y+2\right)=0\)
\(\Rightarrow\left[\begin{matrix}x+3=0\\y+2=0\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=-3\\y=-2\end{matrix}\right.\)
Vậy \(x=-3;y=-2\)
xy + 2x + 3y = -6
=> x ( y + 2 ) + 3y + 6 = 0
=> x ( y + 2 ) + 3 ( y + 2 ) = 0
=> ( x + 3 ) ( y + 2 ) = 0
=> \(\left\{\begin{matrix}x+3=0\\y+2=0\end{matrix}\right.\)=> \(\left\{\begin{matrix}x=-3\\y=-2\end{matrix}\right.\)
Vậy x = -3 , y = -2