Question

Rationalise the denominator of 14/√108-√96+√192-√54​

Answer 1

Answer:

$\boxed{ \huge{ \pink{ \dfrac{2 \sqrt{3} + \sqrt{6} }{3} }}}$

Step-by-step explanation:

$\bold { \underline{ \red{Given}}} \longrightarrow \\ \dfrac{14}{ \sqrt{108} - \sqrt{96} + \sqrt{192} - \sqrt{54} }$

$\bold{ \underline{ \red{To \: find}}} \longrightarrow \\ ratioalization \: of \: given \: fraction$

$\bold{ \underline{\red{Concept \: used}}} \longrightarrow \\ \bold{ \boxed{ \blue{ \large{{x}^{2} - {y}^{2} = (x + y)(x - y)}}}}$

$\bold{ \underline{ \red{Solution}}} \longrightarrow \\ \dfrac{14}{ \sqrt{108} - \sqrt{96} + \sqrt{192} - \sqrt{54} } \\ = \dfrac{14}{ \sqrt{36 \times 3} - \sqrt{16 \times 6} + \sqrt{64 \times 3} - \sqrt{9 \times 6} } \\ = \dfrac{14}{6 \sqrt{3} - 4 \sqrt{6} + 8 \sqrt{3} - 3 \sqrt{6} } \\ = \dfrac{14}{14 \sqrt{3} - 7 \sqrt{6} } \\ = \dfrac{14}{7(2 \sqrt{3} - \sqrt{6} )} \\ = \dfrac{2}{2 \sqrt{3} - \sqrt{3} \sqrt{2} } \\ = \dfrac{2}{ \sqrt{3}(2 - \sqrt{2} ) } \\ = \dfrac{2(2 + \sqrt{2}) }{ \sqrt{3}(2 - \sqrt{2})(2 + \sqrt{2} ) } \\ = \dfrac{2 \sqrt{3} (2 + \sqrt{2} )}{ \sqrt{3} \sqrt{3} \: \: ( {(2)}^{2} - {( \sqrt{2}) }^{2}) } \\ = \dfrac{2(2 \sqrt{3} + \sqrt{6}) }{3(4 - 2)} \\ = \dfrac{2(2 \sqrt{3} + \sqrt{6} ) }{3 ( 2)} \\ = \dfrac{2 \sqrt{3} + \sqrt{6} }{3}$

Answer 2

Answer:

Step-by-step explanation:

miss uhhhhhhhhhhhh Rishabh

Rishu ki malika here !!!!

see in the attachment