A problem by Nazmus Sakib

Algebra Level 1

True or False:

\[\frac{24 + 2 \sqrt{121}}{12 + \sqrt{121}} \in \mathbb{Z}\]

Note: By breaking \(\sqrt,\) if you want to change the sign then the sign of the numerator and denominator should be the same. e.g \(\dfrac{a+\sqrt{b^2}}{c+\sqrt{d^2}} = \dfrac{a+b}{c+d}\) or \(\dfrac{a-b}{c-d}\)

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