$\begin{array}{l}Bài \,\,18:\\ a) \, \sqrt{a^2 + 6a +9}+\sqrt{a^2 - 6a + 9}\\ =\sqrt{(a+3)^2}+\sqrt{(a-3)^2}\\ =|a+3| + |a-3|\\=a+3 - (a - 3)\,\,\,\,[Do\,\,(a - 3 \leq 0)]\\ =6\\ b) \, \sqrt{a + 2\sqrt{a-1}}+\sqrt{a-2\sqrt{a-1}}\\ =\sqrt{a-1 + 2\sqrt{a-1} + 1}+\sqrt{a-1 -2\sqrt{a-1} + 1}\\ =\sqrt{(\sqrt{a-1}+1)^2} + \sqrt{(\sqrt{a-1}-1)^2}\\=|\sqrt{a-1} +1| + |\sqrt{a-1} - 1|\\ =\sqrt{a-1} +1 - (\sqrt{a-1} -1)\,\,\,\,[Do\,\,(\sqrt{a-1} \leq 1)]\\ =2\\ \\ Bài\,\,19:\\ a)\,\dfrac{a\sqrt{a}-8+2a-4\sqrt{a}}{a-4}\\ =\dfrac{(\sqrt{a}-2)(a + 2\sqrt{a} + 4) + 2\sqrt{a}(\sqrt{a} - 2)}{(\sqrt{a} - 2)(\sqrt{a} + 2)}\\=\dfrac{(\sqrt{a}-2)(a + 2\sqrt{a} + 4 + 2\sqrt{a})}{(\sqrt{a} - 2)(\sqrt{a}+2)}\\ =\dfrac{a+4\sqrt{a}+4}{\sqrt{a}+2}\\=\dfrac{(\sqrt{a} + 2)^2}{\sqrt{a} + 2}\\ =\sqrt{a} + 2\\ b) \, \dfrac{12\sqrt{6}}{\sqrt{7+2\sqrt{6}}-\sqrt{7-2\sqrt{6}}}\\ =\dfrac{12\sqrt{6}}{\sqrt{(\sqrt{6}+1)^2}-\sqrt{(\sqrt{6}-1)^2}}\\ =\dfrac{12\sqrt{6}}{\sqrt{6} +1 - (\sqrt{6} - 1)}\\ =6\sqrt{6}\end{array}$
