Bài 3:
\(a,=\dfrac{4\left(\sqrt{3}-1\right)}{2}+\dfrac{\sqrt{3}+2}{-1}+\dfrac{6\left(\sqrt{3}-3\right)}{-6}\\ =2\sqrt{3}-2-\sqrt{3}-2-\sqrt{3}+3=-1\\ b,=\sqrt{\dfrac{\left(\sqrt{3}+\sqrt{2}\right)^2}{1}}+\sqrt{\dfrac{\left(\sqrt{3}-\sqrt{2}\right)^2}{1}}\\ =\dfrac{\sqrt{3}+\sqrt{2}}{1}+\dfrac{\sqrt{3}-\sqrt{2}}{1}=2\sqrt{3}\\ c,=\dfrac{4\left(\sqrt{7}-\sqrt{3}\right)}{4}-\dfrac{2\left(\sqrt{5}+\sqrt{3}\right)}{2}\\ =\sqrt{7}-\sqrt{3}-\sqrt{5}-\sqrt{3}=\sqrt{7}-\sqrt{5}\)
\(d,=\dfrac{\sqrt{ab}-b-a+\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}=\dfrac{-\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{\sqrt{b}-\sqrt{a}}{\sqrt{a}+\sqrt{b}}\)
