Question

1/√5-√2-√7 simplify by rationalizing denominator
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Answer 1

12✓36 please mark me brainleast

Answer 2

Answer:

$\frac{1}{ \sqrt{5} - \sqrt{2} - \sqrt{7} } \\ \frac{1}{ (\sqrt{5} - \sqrt{2} ) - \sqrt{7} } \\ \frac{1}{( \sqrt{5} - \sqrt{2} ) - \sqrt{7} } \times \frac{ (\sqrt{5 } - \sqrt{2} ) + \sqrt{7} }{( \sqrt{5} - \sqrt{2}) + \sqrt{7} } \\ we \: \: will \: \: use \: \: these \: \: identities \: \\ </p><p>( x + y)(x - y) = {x}^{2} - {y}^{2} \\ and \\ (x - y {)}^{2} = {x}^{2} + {y}^{2} - 2xy </p><p> \\ = \frac{( \sqrt{5 } - \sqrt{2}) + \sqrt{7} }{( \sqrt{5} - \sqrt{2} {)}^{2} - ( \sqrt{7} {)}^{2} } \\ = \frac{ \sqrt{5} - \sqrt{2} + \sqrt{7} }{(( \sqrt{5} {)}^{2} + ( \sqrt{2 } {)}^{2} - 2 \times \sqrt{5} \times \sqrt{2}) - 7 } \\ \frac{ \sqrt{5 } - \sqrt{2} + \sqrt{7} }{(5 + 2 - 2 \sqrt{10} ) - 7} \\ = \frac{ \sqrt{5} - \sqrt{2} + \sqrt{7} }{7 - 2 \sqrt{10} - 7 } \\ = \frac{ \sqrt{5} - \sqrt{2} + \sqrt{7} }{ - 2 \sqrt{10} } \\ = \frac{ \sqrt{5} - \sqrt{2} + \sqrt{7} }{ - 2 \sqrt{10} } \times \frac{ - 2 \sqrt{10} }{ - 2 \sqrt{10} } \\ = \frac{ - 2 \sqrt{10}( \sqrt{5} - \sqrt{2} + \sqrt{7} ) }{( - 2 \sqrt{10} {)}^{2} } \\ = \frac{ - 2 \sqrt{50 } + 2 \sqrt{20} - 2 \sqrt{70} }{40} \\ hence \: \: now \: \: denominator \: \: is \: \: rationalized \:$

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