\(\dfrac{4}{\left(\sqrt{3}-1\right)}\)+\(\dfrac{1}{\left(\sqrt{3}-2\right)}\)+\(\dfrac{6}{\left(\sqrt{3}+3\right)}\) = \(\dfrac{4\left(\sqrt{3}+1\right)}{3-1}\) +\(\dfrac{\sqrt{3}+2}{3-4}\) +\(\dfrac{6\left(\sqrt{3}-3\right)}{3-9}\) = \(\dfrac{4(\sqrt{3}+1)}{2}\) _ \(\dfrac{\sqrt{3}+2}{1}\)_ \(\dfrac{6(\sqrt{3}-3)}{6}\)
= \(2\left(\sqrt{3}+1\right)\) - \(\sqrt{3}+2\) - \(\sqrt{3}-3\) = \(2\sqrt{3}+2-\sqrt{3}+2-\sqrt{3}-3\) = 1
CHÚC HỌC TỐT NHA !!
\(\dfrac{4}{\sqrt{3}-1}+\dfrac{1}{\sqrt{3}-2}+\dfrac{6}{\sqrt{3}+3}\)
\(=\dfrac{4\left(\sqrt{3}+1\right)}{3-1}+\dfrac{\sqrt{3}+2}{3-4}+\dfrac{6\left(\sqrt{3}-3\right)}{3-9}\)
\(=2\sqrt{3}+2-\sqrt{3}+2-\sqrt{3}+3\)
\(=7\)
\(\dfrac{4}{\sqrt{3}-1}+\dfrac{1}{\sqrt{3}-2}+\dfrac{6}{\sqrt{3}+3}\\ =\dfrac{4\left(\sqrt{3}+1\right)}{3-1}-\dfrac{\sqrt{3}+2}{4-3}+\dfrac{6\left(3-\sqrt{3}\right)}{9-3}\\ =2\left(\sqrt{3}+1\right)-\left(\sqrt{3}+2\right)+3-\sqrt{3}\\ =2\sqrt{3}+2-\sqrt{3}-2+3-\sqrt{3}\\ =3\)
