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\(pt\left(2\right)\Leftrightarrow\sqrt{2\left(x-y\right)^2+10x-6y+12}-\sqrt{y}-\sqrt{x+2}=0\)
\(\Leftrightarrow\sqrt{2\left(x-y\right)^2+10x-6y+12}-2\sqrt{y}-\left(\sqrt{x+2}-\sqrt{y}\right)=0\)
\(\Leftrightarrow\dfrac{2\left(x-y\right)^2+10x-6y+12-4y}{\sqrt{2\left(x-y\right)^2+10x-6y+12}+2\sqrt{y}}-\dfrac{x+2-y}{\sqrt{x+2}+\sqrt{y}}=0\)
\(\Leftrightarrow\dfrac{2\left(x-y+3\right)\left(x-y+2\right)}{\sqrt{2\left(x-y\right)^2+10x-6y+12}+2\sqrt{y}}-\dfrac{x+2-y}{\sqrt{x+2}+\sqrt{y}}=0\)
\(\Leftrightarrow\left(x-y+2\right)\left(\dfrac{2\left(x-y+3\right)}{\sqrt{2\left(x-y\right)^2+10x-6y+12}+2\sqrt{y}}-\dfrac{1}{\sqrt{x+2}+\sqrt{y}}\right)=0\)
\(\Rightarrow x=y-2\). Thay vào \(pt(1)\) có:
\(pt\left(1\right)\Leftrightarrow\sqrt{y^2-8\left(y-2\right)+9}-\sqrt[3]{\left(y-2\right)y+12-6\left(y-2\right)}\le1\)
\(\Leftrightarrow\sqrt{y^2-8y+25}-\sqrt[3]{y^2-8y+24}\le1\)
\(\Leftrightarrow\left(\sqrt{y^2-8y+25}-3\right)-\left(\sqrt[3]{y^2-8y+24}-2\right)\le0\)
\(\Leftrightarrow\dfrac{y^2-8y+25-9}{\sqrt{y^2-8y+25}+3}-\dfrac{y^2-8y+24-8}{\sqrt[3]{\left(y^2-8y+24\right)^2}+4+2\sqrt[3]{y^2-8y+24}}\le0\)
\(\Leftrightarrow\dfrac{\left(y-4\right)^2}{\sqrt{y^2-8y+25}+3}-\dfrac{\left(y-4\right)^2}{\sqrt[3]{\left(y^2-8y+24\right)^2}+4+2\sqrt[3]{y^2-8y+24}}\le0\)
\(\Leftrightarrow\left(y-4\right)^2\left(\dfrac{1}{\sqrt{y^2-8y+25}+3}-\dfrac{1}{\sqrt[3]{\left(y^2-8y+24\right)^2}+4+2\sqrt[3]{y^2-8y+24}}\right)\le0\)
\(\Rightarrow y=4\Rightarrow x=y-2=4-2=2\)
Vậy \(x=2;y=4\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left|x+5\right|=x+5\\\left|x-5\right|=5-x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+5>=0\\x-5< =0\end{matrix}\right.\Leftrightarrow-5< =x< =5\)