Simplify $\frac{1+\sqrt{2}}{1+\sqrt{3}} \cdot 2$. Your solution can be converted to the form $-A-\sqrt{B}+\sqrt{C}+\sqrt{D}$, where A, B, C, and D are positive integers. What is $A+B+C+D$?
+128474 \( $\frac{1+\sqrt{2}}{1+\sqrt{3}} \cdot 2$\)
\($-A-\sqrt{B}+\sqrt{C}+\sqrt{D}$\)
Multiply num / den by 1 - sqrt 3
(1 + sqrt 2) ( 1 -sqrt 3) 1 -sqrt 2 + sqrt 3 - sqrt 6 1 - sqrt 2 + sqrt 3 - sqrt 6
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( 1 + sqrt 3) (1 - sqrt 3) 1 - 3 -2
Multiplying this result by 2 we get
- 1 ( 1 - sqrt 2 + sqrt 3 - sqrt 6 ) =
-1 - sqrt 3 + sqrt 2 + sqrt 6
A + B + C + D = 1 + 3 + 2 + 6 = 12
+2666 We have: \({1 + \sqrt2 \over 1 + \sqrt 3 } \times 2\)
First, we need to rationalize the denominator of the fraction by multiplying it by \({1 - \sqrt 3 \over 1 - \sqrt 3 } \), which gives us \({\sqrt 2 - \sqrt 6 +1 - \sqrt 3 \over -2}\)
MUltiplying this fraction by 2 gives us: \(-1(\sqrt 2 - \sqrt 6 + 1 - \sqrt 3) = -1 - \sqrt 2 + \sqrt 3 + \sqrt 6 \)
Thus, the sum is \(1 + 2 + 3 + 6 = \color{brown}\boxed{12}\)
