Question

tính

\(\dfrac{2}{\sqrt{3}+1}-\dfrac{1}{\sqrt{3}-2}+\dfrac{12}{\sqrt{3}+3}\)

giúp mk vs ạ mk cần gấp

Answer 1

BT=\(\dfrac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}+\dfrac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+\dfrac{12\left(3-\sqrt{3}\right)}{\left(\sqrt{3}+3\right)\left(3-\sqrt{3}\right)}\)

\(=\dfrac{2\left(\sqrt{3}-1\right)}{2}+\dfrac{2+\sqrt{3}}{4-3}+\dfrac{12\left(3-\sqrt{3}\right)}{9-3}\)

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

\(=\sqrt{3}-1+2+\sqrt{3}+6-2\sqrt{3}=7\)