# Extra Questions For Class 9 Maths Chapter 1 | Q10

Question 10: Rationalize the denominator of $$frac{1}{9 + {\sqrt{5} + \sqrt{6}}}\\$ ## Target Centum in CBSE 10th Maths #### Free Online CBSE Coaching online.maxtute.com ### Video Explanation ## NCERT Solution to Class 10 Maths #### With Videos ### Explanatory Answer | Number Systems $\frac{1}{9 + \sqrt{5} + \sqrt{6}}$ Rationalising the denominator of this question has to be done in two stages. In each of the stages, we will be multiplying the numerator and the denominator of the fraction with the conjugate of the denominator. Stage 1: The conjucate of$9 + $$sqrt{5}$ + $\sqrt{6}$) is$9 - ($$sqrt{5}$ + $\sqrt{6}$)) = $\frac{1}{9 + \sqrt{5} + \sqrt{6}}$ x $\frac{9 -$$sqrt{5} + \sqrt{6}$}{9 -$$sqrt{5} + $sqrt{6}$}) = $\frac{9 -$$sqrt{5} + \sqrt{6}$}{9^2 -$$sqrt{5} + $sqrt{6}$^2}) = $\frac{9 -$$sqrt{5} + \sqrt{6}$}{81 -$5 + 6 + 2$sqrt{30})}) = $\frac{9 -${$sqrt{5} + \sqrt{6}}$}{70 - 2{$sqrt{30}}}\\$ Stage 2: The conjucate of$70 - 2$$sqrt{30}$) is$70 + 2$$sqrt{30}$) $\frac{9 -$\sqrt{5} + \sqrt{6}$}{70 - 2$sqrt{30}}$ x $\frac{70 + 2\sqrt{30}}{70 + 2\sqrt{30}}$
= $\frac{630 + 18\sqrt{30} - 70\sqrt{5} - 10\sqrt{6} - 70\sqrt{6} - 12\sqrt{5}} {4900 - 120}\\$
= $\frac{630 + 18\sqrt{30} - 82\sqrt{5} - 80\sqrt{6}} {4780}$

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