Question 33473
{{{(x^2-3x-10)/(x^2-x-20)}}}
{{{(x^2-3x-10)}}} = {{{(x^2-5x+2x-10)}}}
x(x-5)+2(x-5)
(x+2)(x-5)
{{{(x^2-x-20)}}} = {{{(x^2-5x+4x-20)}}}
x(x-5)+4(x-5)
(x+4)(x-5)
now {{{(x+2)(x-5)/(x+4)(x-5)}}}
{{{(x+2)/(x+4)}}}
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2x^2 + 5x - 12              2x^2 - 7x + 6
__________________    /   ________________
9x^2 - 16                  3x^2-x-4

Now taking the numerator and denominator separately and factorizing seperately

2x^2 + 5x - 12            
__________________  
9x^2 - 16                  



2x^2+8x-3x-12                
-------------           
(3x)^2 - (4)^2                 



2x(x+4)-3(x+4)
--------------
(3x-4)(3x+4) 



(2x-3)(x+4)
 ------------
(3x-4)(3x+4)



Now the denominator


 2x^2 - 7x + 6
__________________    
 3x^2-x-4



2x^2-4x-3x+6               
-------------           
(3x)^2 -4x+3x-4                 



2x(x-2)-3(x-2)
--------------
x(3x-4)+1(3x-4) 



(2x-3)(x+4)
 ------------
(x+1)(3x-4)



Numerator / Denominator



(2x-3)(x+4)
 ------------
(3x-4)(3x+4)
__________________
(2x-3)(x+4)
 ------------
(x+1)(3x-4)



(2x-3)(x+4)     (x+1)(3x-4)
 ------------ * ------------
(3x-4)(3x+4)    (2x-3)(x+4)



cancelling the common terms and simplifying we get

(x+1)
-----
(3x+4)
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Rationalize the denominator

{{{2/(sqrt(6) - sqrt(5))}}}

multiply the numerator and denominator with {{{(sqrt(6) + sqrt(5))}}}

{{{2*(sqrt(6) +sqrt(5))}}}/(6 - 5)
{{{2*(sqrt(6) +sqrt(5))}}}