ĐKXĐ: \(x\ge3\)
Đặt \(\left\{{}\begin{matrix}\sqrt[]{x-3}=a\ge0\\\sqrt[3]{x+4}=b\end{matrix}\right.\) ta được hệ:
\(\left\{{}\begin{matrix}a+b=3\\b^3-a^2=7\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}b=3-a\\b^3-a^2=7\end{matrix}\right.\)
\(\Rightarrow\left(3-a\right)^3-a^2=7\)
\(\Leftrightarrow\left(a-1\right)\left(a^2-7a+20\right)=0\)
\(\Leftrightarrow...\)
`sqrt{x-3}+root{3}{x+4}=3(x>=3)`
`<=>sqrt{x-3}-1+root{3}{x+4}-2=0`
`<=>(x-3-1)/(sqrt{x-3}+1)+(x+4-8)/(root{3}{(x+4)^2}+2root{3}{x+4}+4)=0`
`<=>(x-4)/(sqrt{x-3}+1)+(x-4)/(root{3}{(x+4)^2}+2root{3}{x+4}+4)=0`
`<=>(x-4)(1/(sqrt{x-3}+1)+1/(root{3}{(x+4)^2}+2root{3}{x+4}+4))=0`
Mà `1/(sqrt{x-3}+1)+1/(root{3}{(x+4)^2}+2root{3}{x+4}+4)>0AAx>=3`
`<=>x-4=0<=>x=4(tmdk)`
`->S={4}`