Please solve this? question in the attachment

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2 Answers
May 10, 2018

#=203/9sqrt3#

Explanation:

We will use the fact that #sqrt(a/b)=sqrta/sqrtb#

#7*sqrt1/sqrt3-7/3*sqrt1/sqrt3+3sqrt147#

#=7/sqrt3-7/(3sqrt3)+3sqrt147#

#=(7sqrt3)/3-(7sqrt3)/9+3sqrt147#

#=(21sqrt3)/9-(7sqrt3)/9+3sqrt147#

#=(14sqrt3)/9+3sqrt147#

I've left the #3sqrt147# up till now, cos its nasty to deal with. However, seeing that we have an expression with #sqrt3#, there's a good chance that 147 will be equal to three times a square number. Some good ol' bus shelter long division (or using a calculator if you're allowed) will show you that #147=3xx49# Since 49 is a square number, we can work on simplifying this more

#=(14sqrt3)/9+3[sqrt(3xx49)]#

#=(14sqrt3)/9+3(7sqrt3)#

#=(14sqrt3)/9+21sqrt3#

#=(14sqrt3)/9+(189sqrt3)/9#

#=203/9sqrt3#

May 10, 2018

=39.06736822

Explanation:

#sqrt(1/3) =sqrt3/3#

#sqrt(147)=sqrt49 xx sqrt3=7sqrt3#

#7sqrt(1/3)-2 1/3sqrt(1/3)+3sqrt(147)#

=#(7sqrt3)/3-(7sqrt3)/9+21sqrt3#

#(14sqrt3)/9+21sqrt3=(203sqrt3)/9#

=39.06736822