x2 + y = y2 + x
<=> x2 - y2 + y - x = 0
<=> (x - y)(x + y) - (x - y) = 0
<=> (x - y)(x + y - 1) = 0
<=> \(\left[{}\begin{matrix}x-y=0\\x+y-1=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=y\left(lo\text{ại}\right)\\x+y=1\left(nh\text{ận}\right)\end{matrix}\right.\)
x + y = 1
<=> (x + y)2 = 12
<=> x2 + 2xy + y2 = 1
<=> x2 + y2 = 1 - 2xy
Thay x2 + y2 = 1 - 2xy vào A, ta có:
\(\dfrac{1-2xy+xy}{xy-1}=\dfrac{1-xy}{xy-1}=-1\)
x2 + y = y2 + x
<=> x2 - y2 + y - x = 0
<=> (x - y)(x + y) - (x - y) = 0
<=> (x - y)(x + y - 1) = 0
Mà x - y \(\ne0\) do x \(\ne y\) nên x + y - 1 = 0
=> x + y = 1
\(A=\dfrac{x^2+y^2+xy}{xy-1}=\dfrac{\left(x+y\right)^2-2xy+xy}{xy-1}=\dfrac{1-xy}{xy-1}\)
\(=-1\)