Question

solve these questions using 1 st and 2 nd algebric identity
I will mark the answer brain liest

Answer 1

sorry for my handwriting
I hope you got your answer

Answer 2

Hello friends!!

Here is your answer :

Ans 1

I do it in your other question.

Ans 2

$a = \frac{2 - \sqrt{5} }{2 + \sqrt{5} }$

$a = \frac{2 - \sqrt{5} }{2 + \sqrt{5} } \times \frac{2 - \sqrt{5} }{2 - \sqrt{5} }$

$a = \frac{(2 - \sqrt{5} )(2 - \sqrt{5}) }{(2 + \sqrt{5}) (2 - \sqrt{5}) }$

$a = \frac{ {(2 - \sqrt{5}) }^{2} }{(2 + \sqrt{5} )(2 - \sqrt{5} )}$

Using identity :

( x - y)² = x² + y² - 2xy

( x + y) ( x - y) = x² - y²

$a = \frac{ {(2)}^{2} + {( \sqrt{5}) }^{2} - 2(2)( \sqrt{5} )}{ {(2)}^{2} - {( \sqrt{5} )}^{2} }$

$a = \frac{4 + 5 - 4 \sqrt{5} }{4 - 5}$

$a = \frac{9 - 4 \sqrt{5} }{ - 1}$

$b = \frac{2 + \sqrt{5} }{2 - \sqrt{5} }$

$b = \frac{2 + \sqrt{5} }{2 - \sqrt{5} } \times \frac{2 + \sqrt{5} }{2 + \sqrt{5} }$

$b = \frac{(2 + \sqrt{5})(2 + \sqrt{5} ) }{(2 - \sqrt{5} )(2 + \sqrt{5} )}$

$b = \frac{ {(2 + \sqrt{5}) }^{2} }{(2 - \sqrt{5} )(2 + \sqrt{5} )}$
Using identity :

( x + y)² = x² + y² + 2xy

( x + y) ( x - y) = x² - y²

$b = \frac{ {(2)}^{2} + { (\sqrt{5}) }^{2} + 2(2)( \sqrt{5} )}{ {(2)}^{2} - {( \sqrt{5}) }^{2} }$

$b = \frac{9 + 4 \sqrt{5} }{ - 1}$

$= > {a}^{2} - {b}^{2}$

$= {( \frac{9 - 4 \sqrt{5} }{ - 1} )}^{2} - { (\frac{9+ 4 \sqrt{5} }{ - 1} ) }^{2}$


$= (\frac{81+ 80- 72 \sqrt{5} }{1} ) - ( \frac{81 + 80 + 72 \sqrt{5} }{1} )$

$= 161- 72 \sqrt{5} - 161 - 72\sqrt{5}$

$= - 144 \sqrt{5}$

Hope it helps you... ☺️☺️☺️☺️

# Be Brainly