\(a,=6-\sqrt{15}+\sqrt{15}+5=11\\ b,=\dfrac{\left(\sqrt{15}-2\sqrt{3}\right)\left(\sqrt{5}+2\right)}{5-4}-\left(2+\sqrt{3}\right)\\ =5\sqrt{3}+2\sqrt{15}-2\sqrt{15}-4\sqrt{3}-2-\sqrt{3}\\ =-2\)
\(C=\left(\dfrac{a+3\sqrt{a}}{\sqrt{a}+3}-2\right)\cdot\left(\dfrac{a-1}{\sqrt{a}-1}+1\right)\left(a\ge0;a\ne1\right)\\ C=\left(\dfrac{\sqrt{a}\left(\sqrt{a}+3\right)}{\sqrt{a}+3}-2\right)\cdot\left(\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}-1}+1\right)\\ C=\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)=a-4\)
b: Ta có: \(\dfrac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}-\dfrac{1}{2-\sqrt{3}}\)
\(=\sqrt{3}-2-\sqrt{3}\)
=-2
