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}`

tui giải khác không biết phải không =]]

=>4 \(\left(\sqrt{x+1}\right)^2\)-  4 \(\left(\sqrt{1-x}\right)^2\)+(3 - x) = 3\(\left(\sqrt{1-x}\right)^2\)

= >4(x+1) -4(1-x) + (3-x) = 3(1-x)

=>4x +4 -4 +4x +3 -x = 3 - 3x

=>10x = 0

=> x=0 => pt VN