Answer:
[tex]\dfrac{12-3\sqrt{5}-4\sqrt{6}+\sqrt{30}}{11}[/tex]
Step-by-step explanation:
You have written 3 - ((√6)/4) + √5. The only denominator is 4, which is already rational. If you were to use appropriate grouping symbols (parentheses) we might expect to see ...
(3 -√6)/(4 +√5)
The denominator is rationalized by taking advantage of the factoring of the difference of squares:
a² - b² = (a -b)(a +b)
The denominator will be rational if we can square the value √5. By choosing a multiplier of (4-√5), we can do that:
[tex]\dfrac{3-\sqrt{6}}{4+\sqrt{5}}=\dfrac{(3-\sqrt{6})(4-\sqrt{5})}{(4+\sqrt{5})(4-\sqrt{5})}\\\\=\dfrac{3\cdot 4-3\sqrt{5}-4\sqrt{6}+\sqrt{30}}{4^2-(\sqrt{5})^2}=\dfrac{12-3\sqrt{5}-4\sqrt{6}+\sqrt{30}}{11}[/tex]
