How do you rationalize the denominator and simplify #(12-2sqrt6)/(8-5sqrt2)#?

#color(blue)("Preamble")#

There is a trick to this. Apart from anything else, mathematicians do not like 'roots' in the denominator. It is considered good practice to 'get rid of them'.

The trick: multiply by its self but change the sign in the middle (for the second variable) .

Why? Consider: #(a-b)(a+b) = a^2-b^2#

Observe that everything ends up being squared. So any square roots disappear.
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#color(blue)("Answering the question")#

#color(green)((12-2sqrt6)/(8-5sqrt2)color(red)(xx1)" "=" "color(green)((12-2sqrt6)/(8-5sqrt2)color(red)(xx(8+5sqrt(2))/(8+5sqrt(2)) )#

#=[(12-2sqrt(6))(8+5sqrt(2))]/ [8^2-(5sqrt(2))^2 ]#

#=(96+60sqrt(2)-64sqrt(6)-10sqrt(12))/(64-50 )#
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Consider #10sqrt(12) -> 10sqrt(2^2xx3) = 20sqrt(3)#
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#=(96+60sqrt(2)-64sqrt(6)-20sqrt(3))/(14 )#

#=(48+30sqrt(2)-32sqrt(6)-5sqrt(12))/7#

#color(blue)("I will leave the rest for you to do")#