Đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\ge0\\\sqrt{y}=b\ge0\end{matrix}\right.\) \(\Rightarrow a+b=1\) \(\Rightarrow\left\{{}\begin{matrix}a;b\in\left[0;1\right]\\b=1-a\end{matrix}\right.\)
\(Q=a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\)
\(Q=1-3a\left(1-a\right)=3a^2-3a+1\)
Xét hàm \(Q=f\left(a\right)=3a^2-3a+1\) trên \(\left[0;1\right]\)
\(f\left(0\right)=1\) ; \(f\left(1\right)=1\) ; \(f\left(\frac{1}{2}\right)=\frac{1}{4}\)
\(\Rightarrow Q_{max}=1\) khi \(\left(a;b\right)=\left(0;1\right);\left(1;0\right)\) hay \(\left(x;y\right)=\left(0;1\right);\left(1;0\right)\)
\(Q_{min}=\frac{1}{4}\) khi \(a=b=\frac{1}{2}\) hay \(x=y=\frac{1}{4}\)
