Cách giải:
$a,x^2-6x+9 \leq 4\sqrt{x^2-6x+6}$
$ĐKXĐ:\left[ \begin{array}{l}x \geq 3+\sqrt{3}\\x \leq 3-\sqrt{3}\end{array} \right.$
$\to x^2-6x+9-4\sqrt{x^2-6x+6} \leq 0$
$\to x^2-6x+6-4\sqrt{x^2-6x+6}+4-1 \leq 0$
$\to (\sqrt{x^2-6x+6}-2)^2 -1 \leq 0$
$\to (\sqrt{x^2-6x+6}-3)(\sqrt{x^2-6x+6}-1) \leq 0$
Vì $\sqrt{x^2-6x+6}-3<\sqrt{x^2-6x+6}-1$
$\to \begin{cases}\sqrt{x^2-6x+6} \geq 1\\\sqrt{x^2-6x+6} \leq 3\\\end{cases}$
$\to \begin{cases}x^2-6x+6 \geq 1\\x^2-6x+6 \leq 9\\\end{cases}$
$\to 1 \leq x^2-6x+6 \leq 9$
$\to 4 \leq x^2+6x+9 \leq 12$
$\to 4 \leq (x+3)^2 \leq 12$
$\to 2 \leq |x+3| \leq 2\sqrt{3}$
$+)|x+3| \geq 2$
$\to \left[ \begin{array}{l}x \geq -1\\x \leq -5\end{array} \right.$
Kết hợp ĐKXĐ ta có:
$\left[ \begin{array}{l}x \geq 3+\sqrt{3}\\x \leq -5\end{array} \right.(*)$
$+)|x+3| \leq 2\sqrt{3}$
$\to \left[ \begin{array}{l}x \leq 2\sqrt{3}-3\\x \geq -2\sqrt{3}-3\end{array} \right.$
Kết hợp (*) ta có:
$\left[ \begin{array}{l}x \geq 3+\sqrt{3}\\x \leq -5\end{array} \right.$
