\(\left\{{}\begin{matrix}x^3-x^2y-2xy+2y^2=0\\2+\sqrt[3]{y^3-14}=x-2\sqrt{x^2-2y-1}\end{matrix}\right.\) điều kiện \(\left(\left\{{}\begin{matrix}y^3>14\\x^2>2y+1\end{matrix}\right.\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2\left(x-y\right)-2y\left(x-y\right)=0\\2+\sqrt[3]{y^3-14}=x-2\sqrt{x^2-2y-1}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)\left(x^2-2y\right)=0\\2+\sqrt[3]{y^3-14}=x-2\sqrt{x^2-2y-1}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=y\\2+\sqrt[3]{y^3-14}=x-2\sqrt{x^2-2y-1}\end{matrix}\right.\) vì( \(x^2-2y-1>0\) nên \(x^2-2y\ne0\))
\(\Leftrightarrow\left\{{}\begin{matrix}x=y\\2+\sqrt[3]{x^3-14}=x-2\sqrt{x^2-2x-1}\end{matrix}\right.\)
\(\Rightarrow\sqrt[3]{x^3-14}=x-2-2\sqrt{x^2-2x-1}\)
vì \(\sqrt{x^2-2x-1}\ge0\forall x\)
\(\Leftrightarrow-2\sqrt{x^2-2x-1}\le0\forall x\)
\(\Leftrightarrow x-2-2\sqrt{x^2-2x-1}\le x-2\forall x\)
\(\Leftrightarrow\sqrt[3]{x^3-14}\le x-2\forall x\)
\(\Leftrightarrow x^3-14\le x^3-6x^2+12x-8\)
\(\Leftrightarrow-6x^2+12x+6\ge0\)
\(\Leftrightarrow x^2-2x-1\le0\)
dấu = xảy ra khi \(x^2-2x-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=y=1+\sqrt{2}\\x=y=1-\sqrt{2}\end{matrix}\right.\)