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Rationalize the denominator of 9/(7-3√3)-(6-4√3)
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is it: (9/x) - y OR: 9/(x-y)
\[\frac{9}{7-3√3}-(6-4√3)\]or: \[\frac{9}{(7-3√3)-(6-4√3)}\]
Oh the second one sorry was trying to understand how to do it
ok, so this the the expression you want to rationalize, correct? \[\frac{9}{(7-3√3)-(6-4√3)}\]
Correct
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$$ \frac{9}{1+\sqrt{3}} $$
ok, first simplify the denominator: \[\frac{9}{(7-3√3)-(6-4√3)}=\frac{9}{7-3√3-6+4√3}=\frac{9}{1+\sqrt{3}}\]
well now you could rationalize ..
you can then remove the radicals from the denominator by noting that:\[a^2-b^2=(a+b)(a-b)\]
Lol thanks to both of you
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glad to help :)
so you can multiply the numerator and denominator by:\[(1-\sqrt{3})\]
I assume you are ok from here on?
Yes thanks again :)
yw
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