Giải:
a) $\sqrt{36}+\sqrt{81}-\sqrt{25}$
= $6+9-5$
= $10$
b) $3\sqrt{48}-2\sqrt{75}+5\sqrt{27}$
= $3.\sqrt{16.3}-2\sqrt{25.3}+5\sqrt{9.3}$
= $3.4\sqrt{3}-2.5\sqrt{3}+5.3\sqrt{3}$
= $12\sqrt{3}-10\sqrt{3}+15\sqrt{3}$
= $17\sqrt{3}$
c) $\sqrt{(1-\sqrt{3})^2}-\sqrt{(1+\sqrt{3})^2}$
= $|1-\sqrt{3}|-|1+\sqrt{3}|$
= $(1-\sqrt{3})-(1+\sqrt{3})$
= $1-\sqrt{3}-1-\sqrt{3}$
= $-2\sqrt{3}$
c) $\frac{1}{\sqrt{5}-2}-\frac{4}{3-\sqrt{5}}$
= $\frac{3-\sqrt{5}}{(\sqrt{5}-2)(3-\sqrt{5})}-\frac{4(\sqrt{5}-2)}{(3-\sqrt{5})(\sqrt{5}-2)}$
= $\frac{(3-\sqrt{5})-(4\sqrt{5}-8)}{(3-\sqrt{5})(\sqrt{5}-2)}$
= $\frac{3-\sqrt{5}-4\sqrt{5}+8}{3\sqrt{5}-6-\sqrt{5}\sqrt{5}+2\sqrt{5}}$
= $\frac{11-5\sqrt{5}}{5\sqrt{5}-11}$
= $-\frac{5\sqrt{5}-11}{5\sqrt{5}-11}$
= $-1$
d) $\frac{\sqrt{2}}{2-\sqrt{5}}-\frac{\sqrt{2}}{2+\sqrt{5}}$
= $\frac{\sqrt{2}.(2+\sqrt{5})}{(2-\sqrt{5})(2+\sqrt{5})}-\frac{\sqrt{2}.(2-\sqrt{5})}{(2+\sqrt{5})(2-\sqrt{5})}$
= $\frac{2\sqrt{2}+\sqrt{2}\sqrt{5}}{(2-\sqrt{5})(2+\sqrt{5})}-\frac{2\sqrt{2}-\sqrt{2}\sqrt{5}}{(2+\sqrt{5})(2-\sqrt{5})}$
= $\frac{(2\sqrt{2}+\sqrt{2}\sqrt{5})-(2\sqrt{2}-\sqrt{2}\sqrt{5})}{(2-\sqrt{5})(2+\sqrt{5})}$
= $\frac{2\sqrt{2}+\sqrt{2}\sqrt{5}-2\sqrt{2}+\sqrt{2}\sqrt{5}}{2^2-\sqrt{5}^2}$
= $\frac{2\sqrt{2}\sqrt{5}}{2^2-\sqrt{5}^2}$
= $\frac{2\sqrt{10}}{4-5}$
= $\frac{2\sqrt{10}}{-1}$
= $-2\sqrt{10}$
e) $(18\sqrt{\frac{1}{3}}-\sqrt{48}):\frac{\sqrt{3}-\sqrt{6}}{1-\sqrt{2}}$
= $(2.9\sqrt{\frac{1}{3}}-\sqrt{48})\frac{1-\sqrt{2}}{\sqrt{3}-\sqrt{6}}$
= $(18\sqrt{\frac{1}{3}}-\sqrt{48})\frac{1-\sqrt{2}}{\sqrt{3}-\sqrt{6}}$
= $(3.6\sqrt{\frac{1}{3}}-\sqrt{48})\frac{1-\sqrt{2}}{\sqrt{3}-\sqrt{6}}$
= $(3.\sqrt{6^2}\sqrt{\frac{1}{3}}-\sqrt{48})\frac{1-\sqrt{2}}{\sqrt{3}-\sqrt{6}}$
= $(3\sqrt{6^2.\frac{1}{3}}-\sqrt{48})\frac{1-\sqrt{2}}{\sqrt{3}-\sqrt{6}}$
= $(3\sqrt{\frac{36}{3}}-\sqrt{48})\frac{1-\sqrt{2}}{\sqrt{3}-\sqrt{6}}$
= $(3\sqrt{12}-\sqrt{48})\frac{1-\sqrt{2}}{\sqrt{3}-\sqrt{6}}$
= $(3\sqrt{2^2.3}-\sqrt{16.3})\frac{1-\sqrt{2}}{\sqrt{3}-\sqrt{2.3}}$
= $(6\sqrt{3}-4\sqrt{3})\frac{1-\sqrt{2}}{\sqrt{3}(1-\sqrt{2})}$
= $2\sqrt{3}\frac{1}{\sqrt{3}}$
= $\frac{2\sqrt{3}}{\sqrt{3}}$
= $2$
f) $(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}):\frac{1}{\sqrt{7}-\sqrt{5}}$
= $(\frac{\sqrt{7.2}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{5.3}-\sqrt{5}}{1-\sqrt{3}})\frac{\sqrt{7}-\sqrt{5}}{1}$
= $-(-\frac{\sqrt{7}(\sqrt{2}-1)}{1-\sqrt{2}}-\frac{\sqrt{5}(\sqrt{3}-1)}{1-\sqrt{3}})(\sqrt{7}-\sqrt{5})$
= $-(\frac{\sqrt{7}(1-\sqrt{2})}{1-\sqrt{2}}+\frac{\sqrt{5}(1-\sqrt{3})}{1-\sqrt{3}})(\sqrt{7}-\sqrt{5})$
= $-(\sqrt{7}+\sqrt{5})(\sqrt{7}-\sqrt{5})$
= $-(\sqrt{7}^2-\sqrt{5}^2)$
= $-(7-5)$
= $-2$
g) $(\frac{\sqrt{3}-\sqrt{6}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}):\frac{1}{\sqrt{3}+\sqrt{5}}$
= $(\frac{\sqrt{7.2}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{5.3}-\sqrt{5}}{1-\sqrt{3}})\frac{\sqrt{7}+\sqrt{5}}{1}$
= $(\frac{\sqrt{3}-\sqrt{3.2}}{1-\sqrt{2}}+\frac{\sqrt{5.3}-\sqrt{5}}{1-\sqrt{3}})({\sqrt{3}+\sqrt{5}})$
= $(\frac{\sqrt{3}(1-\sqrt{2})}{1-\sqrt{2}}+\frac{\sqrt{5}(\sqrt{3}-1)}{1-\sqrt{3}})({\sqrt{3}+\sqrt{5}})$
= $(\frac{\sqrt{3}(1-\sqrt{2})}{1-\sqrt{2}}-\frac{-\sqrt{5}(\sqrt{3}-1)}{1-\sqrt{3}})({\sqrt{3}+\sqrt{5}})$
= $(\frac{\sqrt{3}(1-\sqrt{2})}{1-\sqrt{2}}-\frac{-\sqrt{5}(1-\sqrt{3})}{1-\sqrt{3}})({\sqrt{3}+\sqrt{5}})$
= $(\sqrt{3}-\sqrt{5})(\sqrt{3}+\sqrt{5})$
= $(\sqrt{3}^2-\sqrt{5}^2)$
= $3-5$
= $-2$
h) $\frac{\sqrt{x^2-8x+16}}{2+\sqrt{x}}(x≥4)$
= $\frac{\sqrt{(x-4)^2}}{2+\sqrt{x}}$
= $\frac{{|x-4|}}{2+\sqrt{x}}(x≥4)$
= $\frac{{x-4}}{2+\sqrt{x}}$
= $\frac{{\sqrt{x}^2-\sqrt{4}^2}}{2+\sqrt{x}}$
= $\frac{{(\sqrt{x}-2)(\sqrt{x}+2)}}{2+\sqrt{x}}$
= \sqrt{x}-2
i) $\sqrt{(2-\sqrt{3})^2}+\sqrt{4-2\sqrt{3}}$
= $\sqrt{(2-\sqrt{3})^2}+\sqrt{(1-\sqrt{3})^2}$
= $|2-\sqrt{3}|+|1-\sqrt{3}|$
= $2-\sqrt{3}+\sqrt{3}-1$
= 1
k) $\sqrt{8+2\sqrt{15}}-\sqrt{8-2\sqrt{15}}$
= $\sqrt{(\sqrt{5}+\sqrt{3})^2}-\sqrt{(\sqrt{5}-\sqrt{3})^2}$
= $|\sqrt{5}+\sqrt{3}|-|\sqrt{5}-\sqrt{3}|$
= $(\sqrt{5}+\sqrt{3})-(\sqrt{5}-\sqrt{3})$
= $\sqrt{5}+\sqrt{3}-\sqrt{5}+\sqrt{3}$
= $2\sqrt{3}$
