Question

Simplify: $\frac{1}{\sqrt{5}-\sqrt{7}+\sqrt{12}}$

Answer 1

Make use of the difference of squares identity,

$a^2-b^2=(a-b)(a+b)$

Let $a=\sqrt5-\sqrt7$ and $b=\sqrt{12}$. Then

$\dfrac1{(\sqrt5-\sqrt7)+\sqrt{12}}\cdot\dfrac{(\sqrt5-\sqrt7)-\sqrt{12}}{(\sqrt5-\sqrt7)-\sqrt{12}}=\dfrac{(\sqrt5-\sqrt7)-\sqrt{12}}{(\sqrt5-\sqrt7)^2-(\sqrt{12})^2}=\dfrac{\sqrt5-\sqrt7-\sqrt{12}}{5-2\sqrt{35}+7-12}=-\dfrac{\sqrt5-\sqrt7-\sqrt{12}}{2\sqrt{35}}$

Now multiply the numerator and denominator by √(35):

$-\dfrac{\sqrt5-\sqrt7-\sqrt{12}}{2\sqrt{35}}\cdot\dfrac{\sqrt{35}}{\sqrt{35}}=-\dfrac{(\sqrt5-\sqrt7-\sqrt{12})\sqrt{35}}{70}=-\dfrac{5\sqrt7-7\sqrt5-2\sqrt{105}}{70}$

Answer 2

Answer:

Multiply the numerator and denominator by the conjugate.

Exact Form:

1√5−√7+2√3

Decimal Form:

0.75267290…