Question

express in simplest form with a rational denominator.
\\(\dfrac{6}{\sqrt{63}}\\)

Explanation:

Step1: Simplify the square root in the denominator

First, factorize 63. We know that \(63 = 9\times7\), and \(\sqrt{63}=\sqrt{9\times7}\). Using the property of square roots \(\sqrt{ab}=\sqrt{a}\times\sqrt{b}\) (where \(a = 9\) and \(b = 7\) and \(a\) is a perfect square), we get \(\sqrt{9\times7}=\sqrt{9}\times\sqrt{7}=3\sqrt{7}\). So the expression becomes \(\frac{6}{3\sqrt{7}}\).

Step2: Rationalize the denominator

To rationalize the denominator, we multiply the numerator and the denominator by \(\sqrt{7}\) (because the denominator has \(\sqrt{7}\), and multiplying by \(\sqrt{7}\) will make the denominator a rational number).

So, \(\frac{6}{3\sqrt{7}}\times\frac{\sqrt{7}}{\sqrt{7}}=\frac{6\sqrt{7}}{3\times7}\) (since \(\sqrt{7}\times\sqrt{7}=7\)).

Step3: Simplify the fraction

Simplify the numerator and the denominator. First, simplify the constants: \(6\) and \(3\) have a common factor of \(3\), so \(\frac{6\sqrt{7}}{21}=\frac{2\sqrt{7}}{7}\) (dividing numerator and denominator by 3).

Answer:

\(\frac{2\sqrt{7}}{7}\)