\(\sqrt[3]{x+10}+\sqrt[3]{17-x}=3\Leftrightarrow\sqrt[3]{x+10}+\sqrt[3]{27-\left(x+10\right)}=3\)
Đặt \(\sqrt[3]{x+10}=T\Rightarrow x+10=T^3\); khi đó PTTT:
\(T+\sqrt[3]{27-T^3}=3\Leftrightarrow\sqrt[3]{27-T^3}=3-T\)
\(\Leftrightarrow27-T^3=27-27T+9T^2-T^3\)
\(\Leftrightarrow9T^2-27T=0\Leftrightarrow9T\left(T-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}T=0\\T=3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\sqrt[3]{x+10}=0\\\sqrt[3]{x+10}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-10\\x=17\end{matrix}\right.\)
Vậy.............
