Lời giải:
Ta có: \(x^2+y=y^2+x\)
\(\Leftrightarrow x^2+y-y^2-x=0\)
\(\Leftrightarrow (x^2-y^2)-(x-y)=0\)
\(\Leftrightarrow (x-y)(x+y-1)=0\)
Vì \(x\neq y\Rightarrow x-y\neq 0\). Do đó \(x+y-1=0\Leftrightarrow x+y=1\)
Khi đó:
\(A=\frac{x^2+y^2+xy}{xy-1}=\frac{(x+y)^2-2xy+xy}{xy-1}=\frac{1-2xy+xy}{xy-1}\)
\(=\frac{1-xy}{xy-1}=-1\)
Vậy \(A=-1\)
Ta có:\(x^2+y=y^2+x\)
\(\Leftrightarrow x^2-y^2+y-x=0\)
\(\Leftrightarrow\left(x-y\right)\left(x+y\right)-\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(x+y-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-y=0\left(loai\right)\\x+y-1=0\end{matrix}\right.\)
\(\Leftrightarrow x+y=1\)
\(\Rightarrow A=\dfrac{x^2+y^2+xy}{xy-1}\)
\(\Leftrightarrow A=\dfrac{x^2+y^2+2xy-xy}{xy-1}\)
\(\Leftrightarrow A=\dfrac{1-xy}{xy-1}=-1\)
\(x^2+y=y^2+x\Leftrightarrow\left(x^2-y^2\right)=x-y\)
x khác y => x+y =1 ; \(x^2+y^2=2x^2-2x+1\)
\(A=x^2+y^2+\dfrac{xy}{xy}-1\) \(\left\{{}\begin{matrix}x;y\ne0\\A=x^2+y^2=2x^2-2x+1=2y^2-2y+1\end{matrix}\right.\)
A là hàm phụ thuộc Biến x ; hoặc y
