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Rationalise the dinominator root6 ÷root 2+root 3 |
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Answer» Identity USED :(a + b)(a - b) = {a}^{2} - {b}^{2}(a+b)(a−b)=a2−b2Now,\BEGIN{gathered} \frac{ \SQRT{6} }{ \sqrt{2} + \sqrt{3} } \\ \end{gathered}2+36On rationalizing the denominator we get,\begin{gathered} = \frac{ \sqrt{6} }{ \sqrt{2} + \sqrt{3} } \times \frac{ \sqrt{2} - \sqrt{3} }{ \sqrt{2} - \sqrt{3} } \\ \\ = \frac{ \sqrt{6} ( \sqrt{2} - \sqrt{3} )}{ {( \sqrt{2}) }^{2} - {( \sqrt{3} )}^{2} } \\ \\ = \frac{ \sqrt{12} + \sqrt{18} }{2 - 3} \\ \\ = \frac{ \sqrt{2 \times 2 \times 3} + \sqrt{3 \times 3 \times 2} }{ - 1} \\ \\ = - ( \sqrt{ {2}^{2} \times 3} + \sqrt{ {3}^{2} \times 2 } ) \\ \\ = - (2 \sqrt{3} + 3 \sqrt{2} ) \\ \\ = - 2 \sqrt{3} - 3 \sqrt{2} \end{gathered}=2+36×2−32−3=(2)2−(3)26(2−3)=2−312+18=−12×2×3+3×3×2=−(22×3+32×2)=−(23+32)=−23−32 |
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