Question

Prime factorise & rationalise denominator: 14/√108-√96+√192-√54​

Answer 1

Step-by-step explanation:

$\frac{14}{ \sqrt{108} - \sqrt{96} + \sqrt{192} - \sqrt{54} }$

$\sqrt{108} = \sqrt{36 \times 3 } = 6 \sqrt{3} \\ \sqrt{96} = \sqrt{16 \times 6} = 4 \sqrt{6} \\ \sqrt{192} = \sqrt{64 \times 3} = 8 \sqrt{3} \\ \sqrt{54} = \sqrt{9 \times 6} = 3 \sqrt{6}$

$\frac{14}{6 \sqrt{3} - 4 \sqrt{6} + 8 \sqrt{3} - 3 \sqrt{6} }$

$\frac{14}{14 \sqrt{3} - 7 \sqrt{6} }$

$\frac{14}{7(2 \sqrt{3} - \sqrt{6}) }$

$\frac{2}{2 \sqrt{3} - \sqrt{6} }$

$\frac{2}{ \sqrt{2}( \sqrt{2} \times \sqrt{3} - \sqrt{3}) }$

$\frac{ \sqrt{2} }{ \sqrt{6} - \sqrt{3} }$

multiplying with √6+√3 we get

$\frac{ \sqrt{2}( \sqrt{6} + \sqrt{3}) }{( \sqrt{6} + \sqrt{3})( \sqrt{6} - \sqrt{ 3} ) }$

$\frac{ \sqrt{12} + \sqrt{6} }{6 - 3}$

$\frac{2 \sqrt{3} + \sqrt{6} }{3}$

hence this is the final answer