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# Algebra

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If we write sqrt(2) + sqrt(3) + 1/(2*sqrt(2) + 3*sqrt(3)) in the form (a*sqrt(2) + b*sqrt(3))/c such that a, b, and c are positive integers and c is as small as possible, then what is a + b + c?

Jan 30, 2022

#1
+514
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The original expression is:

$$\sqrt{2} + \sqrt{3} + {1\over2\sqrt{2} + 3\sqrt{3}}$$

Next we can simplify the fraction part of the expression by multiplying the denominator by the difference of squares;

The expression is now:

$$\sqrt{2}+\sqrt{3} + {3\sqrt{3}-2\sqrt{2}\over19}$$

Then we can multiply the first two terms of the expression by $$19$$ to get:

$${19\sqrt{2}\over19} + {19\sqrt{3}\over19} + {3\sqrt{3}-2\sqrt{2}\over19}$$

Then adding up the numerators and simplifying we get:

$$17\sqrt{2} + 22\sqrt{3}\over19$$

Thus, $$a$$$$b$$, and $$c$$ are $$17$$$$22$$, and $$19$$ respectively.

Since we are looking for the sum of our three "variables", we add $$17+22+19$$ to get:

$$58$$, which is the answer.

:D

Jan 31, 2022
#2
+360
+3

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

$$\sqrt{2}+\sqrt{3}+\frac{1}{2\sqrt{2}+3\sqrt{3}}$$

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

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

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

$$\frac{14+5\sqrt{6}}{2\sqrt{2}+3\sqrt{3}}$$

$$-\frac{-17\sqrt{2}-22\sqrt{3}}{19}$$

$$\frac{17\sqrt{2}+22\sqrt{3}}{19}$$

17+19+22=58, just the same as proyaop's answer!

Jan 31, 2022