\(\frac{\sqrt{2+\sqrt{3}}}{\sqrt{2}}=\frac{\sqrt{2}.\sqrt{2+\sqrt{3}}}{\sqrt{2}.\sqrt{2}}=\frac{\sqrt{4+2.\sqrt{3}}}{2}=\frac{\sqrt{3+2.\sqrt{3}+1}}{2}\)

\(=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{2}=\frac{\sqrt{3}+1}{2}\)

\(\Leftrightarrow\hept{\begin{cases}y=2x-7\\x^2-2.x.\left(2x-7\right)=\left(7-2x\right).\left[9.\left(2x-7\right)+8x\right]\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y=2x-7\\x^2-4x^2+14x=\left(7-2x\right).\left(26x-63\right)\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y=2x-7\\-3x^2+14x=182x-441-52x^2+126x\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y=2x-7\\49x^2-294x+441=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}y=2x-7\\49\left(x^2-6x+9\right)=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}y=2x-7\\\left(x-3\right)^2=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=3\\y=-1\end{cases}}\)

\(\hept{\begin{cases}x^2-2xy=-y\left(9y+8x\right)\\2x-y=7\end{cases}}\)(=) \(\hept{\begin{cases}x^2-2xy+9y^2+8xy=0\\2x-y=7\end{cases}}\)

(=)\(\hept{\begin{cases}x^2+6xy+9y^2-2xy+2xy=0\\2x-y=7\end{cases}}\)

(=)\(\hept{\begin{cases}\left(x+3y\right)^2=0\\2x-y=7\end{cases}}\) (=)\(\hept{\begin{cases}x+3y=0\\2x-y=7\end{cases}}\)(=)\(\hept{\begin{cases}2x+6y=0\left(1\right)\\2x-y=7\left(2\right)\end{cases}}\)

Trừ vế theo vế (2) cho (1) ta được :-7y=7 =>y=-1=>x=3. Vậy \(^{x^3+y^3^{ }=\left(-1\right)^3+3^3=26}\)