If\[\displaystyle\frac{\sqrt{600} + \sqrt{150} + 4\sqrt{54}}{6\sqrt{32} - 3\sqrt{50} - \sqrt{72}} = a\sqrt{b},\]where $a$ and $b$ are integers and $b$ is as small as possible, find $a+b.$
Let's simplify: (10√6 + 5√6 + 12√6) / (24√2 - 15√2 - 6√2)
Can you take it from here?
+129930 sqrt 600 = 10sqrt 6
sqrt 150 = 5sqrt 6
4 sqrt 54 = 4*3 * sqrt (6) = 12sqrt 6
So.....the numerator = 27sqrt 6
6sqrt 32 = 6 * 4 sqrt 2 = 24 sqrt 2
-3sqrt 50 = - 5*3 sqrt 2 = - 15 sqrt 2
-sqrt 72 = - 6sqrt (2)
So.....the denominator = 3sqrt 2
So
27sqrt 6
_______ = 9 sqrt ( 6/2) = 3 sqrt 3
3 sqrt 2
+15 $\sqrt{600}=\sqrt{6}\cdot\sqrt{100}=10\sqrt{6}$
$\sqrt{150}=\sqrt{25}\cdot\sqrt{6}=5\sqrt{6}$
$4\sqrt{54}=4{\sqrt{9}\cdot\sqrt{6}}=(4\cdot3)\sqrt{6}=12\sqrt{6}$
$6\sqrt{32}=6(\sqrt{16}\cdot\sqrt{2})=(6\cdot4)\sqrt{2}=24\sqrt{2}$
$3\sqrt{50}=3(\sqrt{25}\cdot\sqrt{2})=(3*5)\sqrt{2}=15\sqrt{2}$
$\sqrt{72}=\sqrt{36}\cdot\sqrt{2}=6\sqrt{2}$
$\frac{10\sqrt{6}+5\sqrt{6}+12\sqrt{6}}{24\sqrt{2}-15\sqrt{2}-6\sqrt{2}}=\frac{27\sqrt{6}}{3\sqrt{2}}=\frac{27}{3}\cdot\sqrt{\frac{6}{2}}=9\sqrt{3}$
$9+3=12$
12
