+206 Find $\displaystyle{ \frac{2}{1 + 2\sqrt{3}} + \frac{3}{2 - \sqrt{3}}}$, and write your answer in the form $\displaystyle \frac{A + B\sqrt{3}}{C}$, with the fraction in lowest terms and $A > 0$. What is $A+B+C$?
+1223 \(\frac{2}{1+2\sqrt{3}} \cdot \frac{1-2\sqrt{3}}{1-2\sqrt{3}}+ \frac{3}{2-\sqrt{3}} \cdot \frac{2+\sqrt{3}}{2+\sqrt{3}}\)
\(\frac{2 - 4\sqrt{3}}{13} + \frac{6 + 3\sqrt{3}}{1}\)
\(\frac{(2 - 4\sqrt{3} )+ 13(6 + 3\sqrt{3})}{13}\)
\(\frac{80 +35\sqrt{3}}{13}\)
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+240 Hello oliviapow06...!
\(\frac{2}{1+2\sqrt{3}}+\frac{3}{2-\sqrt{3}}\)
\(=\frac{2\left(2-\sqrt{3}\right)}{\left(1+2\sqrt{3}\right)\left(2-\sqrt{3}\right)}+\frac{3\left(1+2\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(1+2\sqrt{3}\right)}\)
\(=\frac{2\left(2-\sqrt{3}\right)+3\left(1+2\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(1+2\sqrt{3}\right)}\)
\(=\frac{2\left(2-\sqrt{3}\right)+3\left(1+2\sqrt{3}\right)}{3\sqrt{3}-4}\)
\(=\frac{7+4\sqrt{3}}{3\sqrt{3}-4}\)
\(=\frac{37\sqrt{3}+64}{11}\)
\(A=64\)
\(B\sqrt{3}=37\sqrt{3}\)
\(C=11\)
\(64+37\sqrt{3}+11\)
\(=75+37\sqrt{3}\)
