Tìm `x` thỏa mãn: `(\sqrt{x} + 1)² = 6\sqrt{x} - 3 - \sqrt{x - 4}`

`~rai~`

\((\sqrt{x}+1)^2=6\sqrt{x}-3-\sqrt{x-4}\quad(x\ge 4)\\\Leftrightarrow x+2\sqrt{x}+1=6\sqrt{x}-3-\sqrt{x-4}\\\Leftrightarrow x+2\sqrt{x}+1-6\sqrt{x}+3+\sqrt{x-4}=0\\\Leftrightarrow x-4\sqrt{x}+4+\sqrt{x-4}=0\\\Leftrightarrow (\sqrt{x}-2)^2+\sqrt{x-4}=0\\\text{mà }(\sqrt{x}-2)^2\ge 0;\sqrt{x-4}\ge 0\quad\forall x\ge 4\\\Leftrightarrow \begin{cases}(\sqrt{x}-2)^2=0\\\sqrt{x-4}=0\end{cases}\\\Leftrightarrow \begin{cases}\sqrt{x}-2=0\\x-4=0\end{cases}\\\Leftrightarrow \begin{cases}\sqrt{x}=2\\x=4\end{cases}\\\Leftrightarrow \begin{cases}x=4\\x=4\end{cases}\\\Leftrightarrow x=4.\quad\text{(thỏa mãn }x\ge 4)\\\text{Vậy }S=\{4\}.\)