\(\left(a\right)\frac{2\sqrt{15}-2\sqrt{10}+\sqrt{6}-3}{2\sqrt{5}-2\sqrt{10}-\sqrt{3}+\sqrt{6}}\\ =\frac{2\sqrt{5}\left(\sqrt{3}-\sqrt{2}\right)+\sqrt{3}\left(\sqrt{2}-\sqrt{3}\right)}{2\sqrt{5}\left(1-\sqrt{2}\right)-\sqrt{3}+\sqrt{6}}\\ =\frac{2\sqrt{5}\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)}{2\sqrt{5}\left(1-\sqrt{2}\right)-\sqrt{3}\left(1-\sqrt{2}\right)}\\ =\frac{\left(2\sqrt{5}-\sqrt{3}\right)\left(\sqrt{3}-\sqrt{2}\right)}{\left(2\sqrt{5}-\sqrt{3}\right)\left(1-\sqrt{2}\right)}\\ =\frac{\sqrt{3}-\sqrt{2}}{1-\sqrt{2}}\)
\(\left(b\right) \frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\\ =\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+2}\\ =\frac{\sqrt{2}+\sqrt{3}+2+2+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+2}\\ =\frac{\left(\sqrt{2}+\sqrt{3}+2\right)+\left(\sqrt{2}.\sqrt{2}+\sqrt{2}.\sqrt{3}+\sqrt{2}.2\right)}{\sqrt{2}+\sqrt{3}+2}\\=\frac{\left(\sqrt{2}+\sqrt{3}+2\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}{\sqrt{2}+\sqrt{3}+2}\\ =\frac{\left(\sqrt{2}+\sqrt{3}+2\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+2}\\ =1+\sqrt{2}\)
\(\left(c\right)\sqrt{9\left(3-a\right)^2}vớia>3\\ =\sqrt{9}.\sqrt{\left(3-a\right)^2}\\ =3.\left|3-a\right|\\ =-3\left(3-a\right)vì.a>3\\ =3a-9\)
\(\left(d\right)\sqrt{a^2.\left(a-2\right)^2}vớia< 0\\ =\sqrt{\left[a\left(a-2\right)\right]^2}\\ =\left|a\left(a-2\right)\right|=-a.\left[-\left(a-2\right)\right]=a\left(a-2\right)=a^2-2a\)
Chúc bạn học tốt ! 