Vì 2x - 2y - x2 + 2xy - y2 = 0
⇒ 2(x - y) - (x - y)2 = 0
⇒ (x - y) [2 - (x - y)] = 0
⇒ (x - y) (2 - x + y) = 0
⇒ \(\left\{{}\begin{matrix}x-y=0\\2-x+y=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-y=0\\x+y=2\end{matrix}\right.\)
Vì x - y = 0 ⇒ x = y
⇒ x + x = 2
⇒ 2x = 2
⇒ x = 1
Vậy x = y =1
\(2x-2y-x^2+2xy-y^2=0\)
\(\Rightarrow\left(2x-2y\right)+\left(-x^2+2xy-y^2\right)=0\)
\(\Rightarrow2\left(x-y\right)-\left(x^2-2xy+y^2\right)=0\)
\(\Rightarrow2\left(x-y\right)-\left(x-y\right)^2=0\)
\(\Rightarrow\left(x-y\right)\left(x-x+y\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-y=0\\x-x+y=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=y\\x-x+y=0\end{matrix}\right.\)
\(\Rightarrow x-x+x=0\)
\(\Rightarrow x=0\)
Vậy \(x=0\)
2x - 2y - x2 +2xy - y2 = 0
Giải
2x - 2y - x2 +2xy - y2 = 0
2.(x-y)- ( x2 - 2xy + y2 ) = 0
2.( x-y ) + ( x-y )2 = 0
(x-y ). ( 2 - x + y ) = 0
⇒ x - y = 0 hoặc 2 - x + y = 0
x = 0 x + y = 2 - 0
x +y = 2
x = 2 + y
