Question

Simply (3+√3)(2+√2),(3+√3)(3−√3)

Answer 1

Step-by-step explanation:

1.

we have ,

$(3 +\sqrt{3})(2 + \sqrt{2})$

$=>3(2 +\sqrt{2}) +\sqrt{3}(2 + \sqrt{2} )$

$=> 6 + 3 \sqrt{2} + 2 \sqrt{3} + \sqrt{3}. \sqrt{2}$)

$=</strong><strong>></strong><strong> 6 + 3 \sqrt{2} + 2 \sqrt{3} + 6$

2.

Given in question ,

$(3 + \sqrt{3})(3 - \sqrt{3})$

$=> 3(3 - \sqrt{3}) + \sqrt{3}(3 - \sqrt{3})$

=>$9 - 3 \sqrt{3} + 3 \sqrt{3 } - \sqrt{3}. \sqrt{3}$

$=> 9 - 9$

$= 0$

Hope this helps you☺☺

Answer 2

(3+3)(2+2)

=>3(2 +\sqrt{2}) +\sqrt{3}(2 + \sqrt{2} )=>3(2+2)+3(2+2)

=> 6 + 3 \sqrt{2} + 2 \sqrt{3} + \sqrt{3}. \sqrt{2}=>6+32+23+3.2 )

=> 6 + 3 \sqrt{2} + 2 \sqrt{3} + 6=>6+32+23+6

2.

Given in question ,

(3 + \sqrt{3})(3 - \sqrt{3})(3+3)(3−3)

=> 3(3 - \sqrt{3}) + \sqrt{3}(3 - \sqrt{3})=>3(3−3)+3(3−3)

=>9 - 3 \sqrt{3} + 3 \sqrt{3 } - \sqrt{3}. \sqrt{3}9−33+33−3.3

=> 9 - 9=>9−9

=0