Question

please dude do this by rationalizing the denominator

please please please help whoever has will be followed by me and would be marked brain list


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Answer 1

Answer:

hey,see the attachment above

Answer 2

Answer:

$\dfrac{2 \sqrt{3} + 2 - 3 \sqrt{2} + \sqrt{6} }{ 4}$

Step-by-step explanation:

Given : $\dfrac{ \sqrt{2} }{ \sqrt{6} - \sqrt{2} } - \dfrac{ \sqrt{3} }{ \sqrt{6} + \sqrt{2} }$

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On further solving, we get

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→ $\dfrac{ \sqrt{2} }{ \sqrt{6} - \sqrt{2} } \times \dfrac{ \sqrt{6} + \sqrt{2} }{ \sqrt{6} + \sqrt{2} } - \dfrac{ \sqrt{3} }{ \sqrt{6} + \sqrt{2} } \times \dfrac{ \sqrt{6} - \sqrt{2} }{ \sqrt{6} - \sqrt{2} }$

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${\boxed{\red{Identity \ : \ (a + b)(a - b) = a^2 - b^2}}}$

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→ $\dfrac{ \sqrt{2} ( \sqrt{6} + \sqrt{2} )}{ {( \sqrt{6} )}^{2} - {( \sqrt{2} )}^{2} } - \dfrac{ \sqrt{3}( \sqrt{6} - \sqrt{2} ) }{ {( \sqrt{6}) }^{2} - {( \sqrt{2}) }^{2} }$

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

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→ $\dfrac{ \sqrt{2}( \sqrt{6} + \sqrt{2} ) }{4} - \dfrac{ \sqrt{3} ( \sqrt{6} - \sqrt{2} ) }{4}$

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→ $\dfrac{ \sqrt{2} ( \sqrt{6} + \sqrt{2} ) - \sqrt{3}( \sqrt{6} - \sqrt{2} )}{4}$

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→ $\dfrac{ \sqrt{2}( \sqrt{6} ) + \sqrt{2} ( \sqrt{2} ) - \sqrt{3} ( \sqrt{6} ) - \sqrt{3} ( - \sqrt{2} )}{4}$

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→ $\dfrac{ \sqrt{2} ( \sqrt{6}) + \sqrt{2} ( \sqrt{2}) - \sqrt{3} ( \sqrt{6}) + \sqrt{3} ( \sqrt{2} )}{4}$

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→ $\dfrac{ \sqrt{2 \times 6} + \sqrt{2 \times 2} - \sqrt{3 \times 6} + \sqrt{3 \times 2} }{4}$

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→ $\dfrac{ \sqrt{2 \times 2 \times 3} + \sqrt{2 \times 2} - \sqrt{3 \times 2 \times 3} + \sqrt{3 \times 2} }{4}$

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→ $\dfrac{2 \sqrt{3} + 2 - 3 \sqrt{2} + \sqrt{6} }{ 4}$