Question

Tính GTBT sau: \(\dfrac{4}{\sqrt{3}+1}-\dfrac{5}{\sqrt{3}-2}+\dfrac{6}{\sqrt{3}-3}\)

Mình đang cần gấp, mai phải nộp bài rồi, ai biết thì giúp mình với! Thanks nhiều!

Answer 1

\(\dfrac{4}{\sqrt{3}+1}-\dfrac{5}{\sqrt{3}-1}+\dfrac{6}{\sqrt{3}-3}\) \(=\dfrac{4\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}-\dfrac{5\left(\sqrt{3}+2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}+\dfrac{6\left(\sqrt{3}+3\right)}{\left(\sqrt{3}+3\right)\left(\sqrt{3}-3\right)}\)

\(=\dfrac{4\left(\sqrt{3}-1\right)}{2}-\dfrac{5\left(\sqrt{3}+2\right)}{-1}+\dfrac{6\left(\sqrt{3}+3\right)}{-6}\)

\(=2\left(\sqrt{3}-1\right)+5\left(\sqrt{3}+2\right)-\left(\sqrt{3}-3\right)\)

\(=2\sqrt{3}-2+5\sqrt{3}+10-\sqrt{3}-3\)

\(=6\sqrt{3}+5\)