Question

please helpp me plzzz what is 8 over the square rott of 6 minus the square root of 3

Answer 1

@narii is this ur query?

Answer 2

no its not

Answer 3

then?

Answer 4

pls draw

Answer 5

@heybhai

Answer 6

\[8/\sqrt{6}-\sqrt{3}=8\times(\sqrt{6}+\sqrt{3})/((\sqrt{6}-\sqrt{3})\times (\sqrt{6}-\sqrt{3}))\]
using the formula (a-b)(a+b)=(a^2-b^2)
hence,\[8\times (\sqrt{6}+\sqrt{3})/((\sqrt{6}^{2})-(\sqrt{3}^{2}))\]
or,\[8\times (\sqrt{6}+\sqrt{3})/(6-3)=8\times (\sqrt{6}+\sqrt{3})/3\]

Answer 7

thanks

Answer 8

can you help me with another one @heybhai

Answer 9

can you help me with another one @heybhai

Answer 10

sure

Answer 11

simplify this