+0  
 
-2
78
7
avatar+204 

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 $A+B+C$?

 Feb 24, 2021
 #1
avatar+939 
0

\(\frac{2}{1+2\sqrt{3}} \cdot \frac{1-2\sqrt{3}}{1-2\sqrt{3}}+ \frac{3}{2-\sqrt{3}} \cdot \frac{2+\sqrt{3}}{2+\sqrt{3}}\)

 

\(\frac{2 - 4\sqrt{3}}{13} + \frac{6 + 3\sqrt{3}}{1}\)

 

\(\frac{(2 - 4\sqrt{3} )+ 13(6 + 3\sqrt{3})}{13}\)

 

\(\frac{80 +35\sqrt{3}}{13}\)

.
 Feb 24, 2021
 #2
avatar+266 
0

Hello oliviapow06...!

\(\frac{2}{1+2\sqrt{3}}+\frac{3}{2-\sqrt{3}}\)

 

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

 

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

 

\(=\frac{2\left(2-\sqrt{3}\right)+3\left(1+2\sqrt{3}\right)}{3\sqrt{3}-4}\)

 

\(=\frac{7+4\sqrt{3}}{3\sqrt{3}-4}\)

 

\(=\frac{37\sqrt{3}+64}{11}\)


\(A=64\)

\(B\sqrt{3}=37\sqrt{3}\)

\(C=11\)

 

\(64+37\sqrt{3}+11\)

\(=75+37\sqrt{3}\)

 

cheekycheekycheeky

 Feb 24, 2021
 #3
avatar+204 
-2

I don't know how but this is apparently wrong

 Feb 24, 2021
 #4
avatar+266 
+1

Oh, both of these answers or just mine?

DewdropDancer  Feb 24, 2021
 #5
avatar+312 
-1

both of them; it was 112

ellapow  Feb 24, 2021
 #7
avatar+266 
+1

Oh, I'm sorry, I must have messed up on a part.

DewdropDancer  Feb 25, 2021
 #6
avatar
0

Bruh I know you are a alt for clairepow STOP cheating on AOPS. This will not be my last warning.

 Feb 24, 2021

65 Online Users

avatar
avatar
avatar