How do you rationalize the denominator: $\frac{3}{\sqrt{3}}$ $3/sqrt3$ ?

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Answer 1 · 2018-03-07T19:36:51

$\sqrt{3}$ $sqrt3$

This means that you do not want an irrational number (surd) in the denominator

$\frac{3}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \leftarrow \frac{\sqrt{3}}{\sqrt{3}} = 1$ $3/sqrt3 xx sqrt3/sqrt3" "larr sqrt3/sqrt3 =1$

$= \frac{3 \sqrt{3}}{{\sqrt{3}}^{2}}$ $=(3sqrt3)/(sqrt3^2)$

$=(cancel3sqrt3)/cancel3$

$=sqrt3$

Answer 2 · 2018-03-07T19:41:12

$3/sqrt3=color(blue)((3sqrt3)/3$

Given:

$sqrt3$

Rationalize the denominator by multiplying the numerator and denominator by $sqrt3$ .

$(3*sqrt3)/(sqrt3sqrt3)$

Apply the rule $sqrtasqrta=a$ .

$(3sqrt3)/3$

Simplify.

$(color(red)cancel(color(black)(3))^1sqrt3)/color(red)cancel(color(black)(3))^1$

$sqrt3$ $larr$ answer

How do you rationalize the denominator: 3/sqrt33√33/sqrt3?