$\text{Bài 1b}$$\\$$\sqrt{5}-$$\dfrac{8}{\sqrt{5}+1}$+$\dfrac{2\sqrt{5}-5}{2-\sqrt{5}}$$\\$$=$$\sqrt{5}-$$\dfrac{8(\sqrt{5}-1)}{(\sqrt{5}+1)(\sqrt{5}-1)}+$$\dfrac{\sqrt{5}(2-\sqrt{5})}{2-\sqrt{5}}$$\\$$=\sqrt{5}-$$\dfrac{8(\sqrt{5}-1)}{5-1}+\sqrt{5}$$\\$$=2\sqrt{5}-$$\dfrac{8(\sqrt{5}-1)}{4}$$\\$$=2\sqrt{5}-2(\sqrt5-1)$$\\$$=2\sqrt5-2\sqrt5+2$$\\$$=2$ $\\$$\text{Bài 2b}$$\\$$2\sqrt{9x-18}-\sqrt{x-2}+\dfrac{1}{2}\sqrt{4x-8}=18$$\\$$2\sqrt{9(x-2)}-\sqrt{x-2}+\dfrac{1}{2}\sqrt{4(x-2)}=18$$\\$$6\sqrt{x-2}-\sqrt{x-2}+\sqrt{x-2}=18$$\\$$6\sqrt{x-2}=18\\\sqrt{x-2}=18:6\\\sqrt{x-2}=3\\\sqrt{x-2}=\sqrt9\\x-2=9\\x=9-2=7\\\text{vậy nghiệm của phương trình đã cho là x=7}$