x+y+xy+1=0
<=>x+xy+y+1=0
<=>x.(y+1)+(y+1)=0
<=>(y+1)(x+1)=0
<=>y+1=0 và x+1=0
<=>y=-1 và x=-1
vậy x=y=-1
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x+y+xy+1=0
<=>x+xy+y+1=0
<=>x.(y+1)+(y+1)=0
<=>(y+1)(x+1)=0
<=>y+1=0 và x+1=0
<=>y=-1 và x=-1
vậy x=y=-1
tim x y biet x+y+xy+1=0
tim x y biet x+y+xy+1=0
tim x y biet x+y+xy+1=0
tim x y biet x+y+xy+1=0
tim x y biet x+y+xy+1=0
tim x y biet x+y+xy+1=0
tim x y biet x+y+xy+1=0
tim xy biet (x-0,2)^10+(y+3,1)=0
bai 1: Tim x biet
\(\hept{\begin{cases}x-y=\frac{3}{10}\\y\left(x-y\right)=-\frac{3}{50}\end{cases}}\)
bai 2: Tim x, y biet:
x+\(\left(-\frac{31}{12}\right)^2\)=\(\left(\frac{49}{12}\right)^2\)-x=y2
Bai 9: Tim x,y,z biet:
(x-1)2+(x+y)2+(xy-z)2=0