$$\large{\left[ \left( \frac{x-3}{x^2-x-6} \right)+\left( \frac{2x-1}{2x^2-5x-3} \right)-\left( \frac{(2x+5)}{x^2+5x+6} \right) \right]}$$is equal to???
@lgbasallote @mathslover @mukushla :)

Question

$$\large{\left[ \left( \frac{x-3}{x^2-x-6} \right)+\left( \frac{2x-1}{2x^2-5x-3} \right)-\left( \frac{(2x+5)}{x^2+5x+6} \right) \right]}$$is equal to???
@lgbasallote @mathslover @mukushla :)

lgbasallote · Answer

lol wow. i suggest factoring first. to make everyone'slives easier

lgbasallote · Answer

$$\frac{x-3}{(x-3)(x+2)} + \frac{2x-1}{(2x +1)(x-3)} - \frac{2x+5}{(x+3)(x+2)}$$

lgbasallote · Answer

there now it's cancelable

lgbasallote · Answer

$$\frac{1}{x+2} + \frac{2x-1}{(2x+1)(x-3)} - \frac{2x+5}{(x+3)(x+2)}$$

lgbasallote · Answer

$$(x+2)(2x+1)(x+3)(x-3)$$
that's the LCD

lgbasallote · Answer

so we have $$\large \frac{(2x+1)(x^2 - 9) +(2x-1)(x+2)(x+3) - (2x+5)(2x+1)(x-3)}{(x+2)(2x+1)(x^2-9)}$$

lgbasallote · Answer

now combine

anonymous · Answer

And your final answer is (For you to verify)
$$
\frac{x-3}{x^2-x-6}-\frac{2
   x+5}{x^2+5 x+6}+\frac{2
   x-1}{2 x^2-5 x-3}=\frac{10
   x}{(x-3) (x+3) (2 x+1)}
$$

jiteshmeghwal9 · Answer

but the answer is given 1 @lgbasallote @eliassaab

mathslover · Answer

should i help ?

jiteshmeghwal9 · Answer

yes, probably.

mathslover · Answer

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