Câu d) ĐK:$x\geq 0$
$x+\sqrt{24+\sqrt{x}}=6$
$\Leftrightarrow \sqrt{24+\sqrt{x}}=6-x$
$\Leftrightarrow \left\{\begin{matrix} 0\leq x\leq 6 & & \\ 24+\sqrt{x}=\left ( 6-x \right )^{2} \left ( 2 \right )& & \end{matrix}\right.$
$\left ( 2 \right )\Leftrightarrow \sqrt{x}-\left ( x^{2}-12x+12 \right )=0$
$\Leftrightarrow \left ( \sqrt{x}-1 \right )-\left ( x^{2}-12x+11 \right )=0$
$\Leftrightarrow \frac{x-1}{\sqrt{x}+1}-\left ( x-1 \right )\left ( x-11 \right )=0$
$\Leftrightarrow \left ( x-1 \right )\left ( \frac{1}{\sqrt{x}+1}-x+11 \right )=0$
$\Leftrightarrow x=1$ hoặc $\frac{1}{\sqrt{x}+1}-x+11=0$
Kết hợp với $x\leq 6$$\Rightarrow \frac{1}{\sqrt{x}+1}-x+11\geq \frac{1}{\sqrt{6}+1}-6+11> 0$$\Rightarrow \frac{1}{\sqrt{x}+1}-x+11=0$ vô nghiệm
Vậy $x=1$