Question

6/sqrt3+2

Answer 1

We have
\[\frac{6}{\sqrt {3}}+2\]
\[6=3 \times 2\]
so
\[\frac{3 \times 2}{\sqrt {3}}+2\]
Also
\[3= \sqrt 3 \times \sqrt 3\]
so
\[\frac{ \sqrt 3 \times \sqrt 3 \times 2}{\sqrt {3}}+2\]
We get finally
\[2\sqrt 3+2\]
or
\[2(1+\sqrt 3)\]

Answer 2

The sqrt of 3+2 is both divided into 6
6/(sqrt3+2)

Answer 3

Is this your question
\[\frac{6}{\sqrt{3}+2}\]

Answer 4

yes

Answer 5

Ok, We have
\[\frac{6}{\sqrt 3+2}\]
Multiply numerator and denominator by the conjugate of denominator \(\sqrt 3-2\)
So we have

\[\frac{6}{\sqrt 3+2} \times \frac{\sqrt 3 -2}{\sqrt 3-2}\]
We get
\[\frac{6 \sqrt 3 -12}{(\sqrt 3)^2-(2)^2}
\]
We get
\[\frac{6 \sqrt 3 -12}{(3-4)}

\]

we get
\[\frac{6 \sqrt 3 -12}{(-1)}\]
Finally we get
\[12-6 \sqrt 3\ or\ 6(2-\sqrt 3)\]

Answer 6

Thank you!

Answer 7

Welcome:D