Lời giải:
\(x^3+y^3+8=6xy\)
\(\Leftrightarrow (x+y)^3-3xy(x+y)+8-6xy=0\)
\(\Leftrightarrow [(x+y)^3+2^3]-3xy(x+y+2)=0\)
\(\Leftrightarrow (x+y+2)[(x+y)^2-2(x+y)+4]-3xy(x+y+2)=0\)
\(\Leftrightarrow (x+y+2)(x^2+y^2+4-xy-2x-2y)=0\)
\(\Rightarrow \left[\begin{matrix} x+y+2=0\\ x^2+y^2+4-xy-2x-2y=0\end{matrix}\right.\)
Nếu $x+y+2=0\Rightarrow x+y=-2$
\(P=4(x+y)-(x+2).\frac{(2+y)}{y}.\frac{y+x}{x}=4(x+y)-\frac{(xy+2x+2y+4)(x+y)}{xy}\)
\(=4(-2)-\frac{[xy+2(-2)+4](-2)}{xy}=-8-(-2)=-6\)
Nếu \(x^2+y^2+4-xy-2x-2y=0\)
\(\Leftrightarrow 2x^2+2y^2+8-2xy-4x-4y=0\)
\(\Leftrightarrow (x^2-2xy+y^2)+(x^2-4x+4)+(y^2-4y+4)=0\)
\(\Leftrightarrow (x-y)^2+(x-2)^2+(y-2)^2=0\)
Từ đây dễ dàng suy ra \((x-y)^2=(x-2)^2=(y-2)^2=0\Rightarrow x=y=2\)
\(P=4(2+2)-(2+2)(\frac{2}{2}+1)(\frac{2}{2}+1)=0\)
