<PRE><B>Question 234575</B>
1/X^2+3x-10 + 1/x-6 all over 1/x-5 - 1/X^2-8x+12 that is my first on=5)(xe
--------------
The lcd of the left side and of the right side are the
same and they cancel leaving:

[(x-6)+(x^2+3x-10] over [x^2-8x+12-(x-5)] 
Simplify:
x^2+4x-16 over x^2-9x+17
======================================

the second one is 5/x + 3/2x all over 1/3x + 1/4x
---
(10x+3x)/2x^2 all over (4x+3x)/12x^2
---
13/2x over 7/12x
Invert the denominator and multiply
(13/2x)(12x/7)
(13*6)/7
78/7
====================
Cheers,
Stan H. 
</PRE>