Question

Rationalise the denominator.;-

√6
____
√3-√2

And simplify

$\sqrt{6}$_______
$\sqrt{3} \: - \sqrt{2}$
So , I want to know that what is the difference between simplify and rationalise the denominator.

Answer 1

Rationalizing the denominator is same the case of above example

Answer 2

HELLO DEAR,


--------------------(1)----------------------

rationalise the denominator means

$\frac{ \sqrt{6} }{ \sqrt{3} - \sqrt{2} } \\ \\ = > \frac{ \sqrt{6} }{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ = > \frac{ \sqrt{6}( \sqrt{3} + \sqrt{2} )}{3 - 2} \\ \\ = > \sqrt{6} ( \sqrt{3} + \sqrt{2} ) \\ \\ = > 3 \sqrt{2} + 2 \sqrt{3}$


take another example for understanding

$\frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} }$
$\frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ = > \frac{3 + 2 + 2 \sqrt{6} }{3 - 2} \\ \\ = > 8 + 2 \sqrt{6}$



--------------------(2)----------------------


and simply MEANS its simple form


$\frac{ \sqrt{6} }{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ = > \frac{ \sqrt{6} ( \sqrt{3} + \sqrt{2} )}{ 3 - 2} \\ \\ = > \sqrt{18} + \sqrt{12} \\ \\ = > 3 \sqrt{2} + 2 \sqrt{3}$


take another example for better understanding

(2x-3)²-8x+12 

=> (2x)²+(3)²-2×(2x)×(3) -8x +12

=> 4x²+9 - 12x - 8x + 12

=> 4x² - 20x +21

=> 4x² - 6x - 14x + 21

=> 2x(2x-3) -7(2x-3)

=> (2x-3) (2x-7)


this is the simple form



nothing much difference in rationalize the denominator and simply



I HOPE ITS HELP YOU DEAR,
THANKS