Question

I missed the class on rationalising denominators.
Can anyone please help me with these three maths along with explanations.

Answer 1

Answer:

$4\sqrt{3}$

$\dfrac{5 - \sqrt{3}}{11}$

$\dfrac{11(3 + \sqrt{5})}{8}$

Step-by-step explanation:

If you have a simple square root in the denominator, multiply the fraction by a fraction that is the root over the root.

$\dfrac{12}{\sqrt{3}} =$

$= \dfrac{12}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}}$

$= \dfrac{12\sqrt{3}}{\sqrt{3}\sqrt{3}}$

$= \dfrac{12\sqrt{3}}{3}$

$= 4\sqrt{3}$

If you have a denominator consisting of a rational number plus a root, multiply the fraction by a fraction that is the denominator over the denominator in which you change only the sign outside the root.

$\dfrac{2}{5 + \sqrt{3}} =$

$= \dfrac{2}{5 + \sqrt{3}} \times \dfrac{5 - \sqrt{3}}{5 - \sqrt{3}}$

$= \dfrac{2(5 - \sqrt{3})}{(5 + \sqrt{3})(5 - \sqrt{3})}$

$= \dfrac{2(5 - \sqrt{3})}{25 - 3}$

$= \dfrac{2(5 - \sqrt{3})}{22}$

$= \dfrac{5 - \sqrt{3}}{11}$

$\dfrac{11}{6 - 2\sqrt{5}} =$

$= \dfrac{11}{6 - 2\sqrt{5}} \times \dfrac{6 + 2\sqrt{5}}{6 + 2\sqrt{5}}$

$= \dfrac{11(6 + 2\sqrt{5})}{(6 - 2\sqrt{5})(6 + 2\sqrt{5})}$

$= \dfrac{22(3 + \sqrt{5})}{36 - 4 \times 5}$

$= \dfrac{22(3 + \sqrt{5})}{36 - 20}$

$= \dfrac{22(3 + \sqrt{5})}{16}$

$= \dfrac{11(3 + \sqrt{5})}{8}$

Answer 2

Answer:

see attached

Step-by-step explanation: