Question

OMG HELP PLEASE!!!!!!!!!!

Answer 1

Answer:

Step-by-step explanation:

Hello, please consider the following.

We will multiply the numerator and denominator by

$3-\sqrt{3}$ to get rid of the root in the denominator.

Let's do it!

$\begin{aligned}\dfrac{\sqrt{12}}{\sqrt{3}+3}&=\dfrac{\sqrt{2^2*3}*(3-\sqrt{3})}{(3+\sqrt{3})*(3-\sqrt{3})}\\\\&=\dfrac{2\sqrt{3}*(3-\sqrt{3})}{3^2-\sqrt{3}^2}\\\\&=\dfrac{2\sqrt{3}*(3-\sqrt{3})}{9-3}\\\\&=\dfrac{2\sqrt{3}*(3-\sqrt{3})}{6}\\\\&=\dfrac{6\sqrt{3}}{6}-\dfrac{2*3}{6}\\\\&=\sqrt{3}-1\\\\\large &\boxed{=-1+\sqrt{3}}\\\end{aligned}$

Thank you

Answer 2

Step-by-step explanation:

Here,

$= \frac{ \sqrt{12} }{ \sqrt{3} + 3}$

is given.

Now, rationalizing it,

$= \frac{ \sqrt{12} }{ \sqrt{3} + 3} \times \frac{ \sqrt{3} - 3 }{ \sqrt{3} - 3}$

now, simplifying it,

$= \frac{ \sqrt{12 } \times \sqrt{3 } - 3 }{( { \sqrt{3} )}^{2} - 9}$

or, simplifying it we get,

$= \frac{3}{ - 6}$

= 1/ -2

Hope it helps...