Question

simplify: (√3+1)(1-√12)+9/√3+√12

Answer 1

Hey friend,
Here is the answer you were looking for:
$\frac{( \sqrt{3} + 1)(1 - \sqrt{12}) + 9 }{ \sqrt{3} + \sqrt{12} } \\ \\ = \frac{( \sqrt{3} \times 1 - \sqrt{3} \times \sqrt{12} + 1 \times 1 - 1 \times \sqrt{12} ) + 9 }{ \sqrt{3} + \sqrt{12} } \\ \\ = \frac{( \sqrt{3} - \sqrt{36} + 1 - \sqrt{12}) + 9 }{ \sqrt{3} + \sqrt{12} } \\ \\ = \frac{ \sqrt{3} - 6 + 12 \sqrt{3} + 9 }{ \sqrt{3} + 2 \sqrt{3} } \\ \\ = \frac{13 \sqrt{3} + 3 }{ \sqrt{3} + 2 \sqrt{3} } \\ \\ on \: rationalizing \: the \: denominator \: we \: get \\ \\ = \frac{13 \sqrt{3} + 3}{ \sqrt{3} + 2 \sqrt{3} } \times \frac{ \sqrt{3} - 2 \sqrt{3} }{ \sqrt{3} - 2 \sqrt{3} } \\ \\ using \: the \: identity \\ (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ = \frac{13 \sqrt{3} \times \sqrt{3} - 13 \sqrt{3} \times 2 \sqrt{3} + 3 \times \sqrt{3} - 3 \times 2 \sqrt{3} }{ {( \sqrt{3}) }^{2} - {(2 \sqrt{3}) }^{2} } \\ \\ = \frac{13 \times 3 - 26 \times 3 + 3 \sqrt{3} - 6 \sqrt{3} }{3 - 12} \\ \\ = \frac{39 - 78 - 3 \sqrt{3} }{ - 9} \\ \\ = \frac{ - 39 - 3 \sqrt{3} }{ - 9} \\ \\ = - ( \frac{ - 39 - 3 \sqrt{3} }{9} ) \\ \\ = \frac{39 + 3 \sqrt{3} }{9} \\ \\ = \frac{13 + \sqrt{3} }{3}$


Hope this helps!!!

@Mahak24

Thanks.....
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