\(\left\{{}\begin{matrix}x+2=a^3\\x+1=b^3\\y-3=c^3\\y-4=d^3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b+c+d=0\\a^3-b^3=1\\c^3-d^3=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+d=-\left(b+c\right)\\a^3+d^3-\left(b^3+c^3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+d=-\left(b+c\right)\\\left(a+d\right)\left(a^2-ad+d^2\right)=\left(b+c\right)\left(b^2-bc+c^2\right)\end{matrix}\right.\) (1)
TH1: \(a+d=-\left(b+c\right)\ne0\)
Chia vế cho vế 2 pt (1) ta được:
\(a^2-ad+d^2=-\left(b^2-bc+c^2\right)\)
\(\Leftrightarrow\left(a-\frac{d}{2}\right)^2+\frac{3d^2}{4}+\left(b-\frac{c}{2}\right)^2+\frac{3c^2}{4}=0\)
\(\Leftrightarrow a=b=c=d=0\) (vô nghiệm)
TH2: \(a+d=-\left(b+c\right)=0\Rightarrow\left\{{}\begin{matrix}a=-d\\b=-c\end{matrix}\right.\)
\(\Rightarrow x+2=4-y\Rightarrow x+y=2\)
\(\Rightarrow A=x^2+y^2\ge\frac{1}{2}\left(x+y\right)^2=2\)
Dấu "=" xảy ra khi \(x=y=1\)