Ta có: $x=y\Leftrightarrow \left\{ \begin{array}{l} x^3+y^3=\frac{(x+y)^3}{4}\\ 2(x^2+y^2)=(x+y)^2 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} 2\sqrt[3]{\frac{x^3+y^3}{2}}=x+y\\ \sqrt{2(x^2+y^2)}:2=(x+y):2 \end{array} \right.\Leftrightarrow \frac{x+y}{2}+\frac{\sqrt{2(x^2+y^2)}}{2}=2\sqrt{\frac{x^3+y^3}{2}}=x+y\Leftrightarrow$ Đ$PCM$
Ta có: $x=y\Leftrightarrow \left\{ \begin{array}{l} x^3+y^3=\frac{(x+y)^3}{4}\\ 2(x^2+y^2)=(x+y)^2 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} 2\sqrt[3]{\frac{x^3+y^3}{2}}=x+y\\ \sqrt{2(x^2+y^2)}:2=(x+y):2 \end{array} \right.\Leftrightarrow \frac{x+y}{2}+\frac{\sqrt{2(x^2+y^2)}}{2}=2\sqrt{\frac{x^3+y^3}{2}}=x+y\Leftrightarrow ĐPCM$
Ta có: $x=y\Leftrightarrow \left\{ \begin{array}{l} x^3+y^3=\frac{(x+y)^3}{4}\\ 2(x^2+y^2)=(x+y)^2 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} 2\sqrt[3]{\frac{x^3+y^3}{2}}=x+y\\ \sqrt{2(x^2+y^2)}:2=(x+y):2 \end{array} \right.\Leftrightarrow \frac{x+y}{2}+\frac{\sqrt{2(x^2+y^2)}}{2}=2\sqrt{\frac{x^3+y^3}{2}}=x+y\Leftrightarrow
$ Đ
$PCM$