Question

Rationalise the denomintror of 0__3-√2-√5

Answer 1

Hey here is your answer

$\frac{0}{3 - \sqrt{2} - \sqrt{5} } \\ \\ \frac{0}{3 - \sqrt{2} - \sqrt{5} } \times \frac{3 - \sqrt{2} + \sqrt{5} }{3 - \sqrt{2} + \sqrt{5}} \\ \\ \frac{0}{(3 - \sqrt{2} )- \sqrt{5} } \times \frac{(3 - \sqrt{2}) + \sqrt{5} }{(3 - \sqrt{2} ) + \sqrt{5}}$

using identity a² - b² = (a + b)(a - b)

$\frac{0(3 - \sqrt{2} + \sqrt{5})}{ {(3 - \sqrt{2} )}^{2} - {( \sqrt{5}) }^{2} }$

using identity (a - b)² = a² + b² - 2ab

$\frac{0(3 - \sqrt{2} + \sqrt{5})}{ {(3 - \sqrt{2} )}^{2} - {( \sqrt{5}) }^{2} } \\ \\ \frac{0}{ {(3)}^{2} + {( \sqrt{2}) }^{2} - 2 \times 3 \times \sqrt{2} - 5 } \\ \\ \frac{0}{9 + 2 - 6 \sqrt{2} - 5} \\ \\ \frac{0}{6 - 6 \sqrt{2} }$

now rationalise :-

$\frac{0}{6 - 6 \sqrt{2} } \\ \\ \frac{0}{6(1 - \sqrt{2}) } \\ \\ \frac{0}{6(1 - \sqrt{2} )} \times \frac{1 + \sqrt{2} }{1 + \sqrt{2} } \\ \\ \frac{0(1 + \sqrt{2}) }{6 \times ( {1}^{2} - { \sqrt{2} }^{2}) } \\ \\ \frac{0}{6(1 - 2)} \\ \\ \frac{0}{6 \times ( - 1)} \\ \\ \frac{0}{ - 6} \\ \\ \bf \: 0 \: \: \: answer \:$
hope this helps you

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