Question #30bd9

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Answer 1 · 2017-11-07T16:26:18

$6 \sqrt{2} - 6 \sqrt{3} + \sqrt{5}$ $6sqrt(2)-6sqrt(3)+sqrt(5)$

$6 \sqrt{2} - 20 \frac{\sqrt{1}}{\sqrt{5}} - \sqrt{108} + \sqrt{125}$ $6sqrt(2)-20sqrt(1)/sqrt(5)-sqrt(108)+sqrt(125)$

First you mus get rid of the sqrt in the numerator:

$6 \sqrt{2} - 4 \cdot 5 \cdot (\frac{1}{\sqrt{5}}) - \sqrt{108} + \sqrt{125}$ $6sqrt(2)-4*5*(1/sqrt(5))-sqrt(108)+sqrt(125)$

$6 \sqrt{2} - 4 \sqrt{5} - \sqrt{108} + \sqrt{125}$ $6sqrt(2)-4sqrt(5)-sqrt(108)+sqrt(125)$

Then decompose the larger number in prime factors:

$6 \sqrt{2} - 4 \sqrt{5} - \sqrt{2^{2} \cdot 3^{3}} + \sqrt{5^{3}}$ $6sqrt(2)-4sqrt(5)-sqrt(2^2*3^3)+sqrt(5^3)$

Knownig that $\sqrt{a^{2}} = a$ $sqrt(a^2)=a$ we can do this:

$6 \sqrt{2} - 4 \sqrt{5} - 2 \cdot 3 \sqrt{3} + 5 \sqrt{5}$ $6sqrt(2)-4sqrt(5)-2*3sqrt(3)+5sqrt(5)$

Simplifing gives:

$6 \sqrt{2} - 6 \sqrt{3} + \sqrt{5}$ $6sqrt(2)-6sqrt(3)+sqrt(5)$