Question

rationalise the denominator 7+3*√5/7-3*√5

Answer 1

Heya !!!

Using the identities :

${(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy \\ (x + y)(x - y) = {x}^{2} - {y}^{2}$


$\frac{7 + 3 \sqrt{5} }{7 - 3 \sqrt{5} }$

On rationalizing the denominator we get,

$= \frac{7 + 3 \sqrt{5} }{7 - 3 \sqrt{5} } \times \frac{7 + 3 \sqrt{5} }{7 + 3 \sqrt{5} } \\ \\ = \frac{ {(7)}^{2} + {(3 \sqrt{5} )}^{2} + 2(7)(3 \sqrt{5} )}{ {(7)}^{2} - {(3 \sqrt{5}) }^{2} } \\ \\ = \frac{49 + 45 + 42 \sqrt{5} }{49 - 45} \\ \\ = \frac{94 + 42 \sqrt{5} }{4} \\ \\ = \frac{47 + 21 \sqrt{5} }{2}$

Hope this helps ☺

Answer 2

Hey !!!!! Your answer =>

Refer to the attached file ^^^

Hope it will help you ☆▪☆ Thanks ^_^

☆ Be Brainly ☆