Question: Simplify: $\dfrac{1}{a²-5a+6} -$$\dfrac{2}{a²-4a+3} -$$\dfrac{1}{a²-3a+2}$
Solution:
Given,
$= \dfrac{1}{a²-5a+6} -\dfrac{2}{a²-4a+3} -\dfrac{1}{a²-3a+2}$
$= \dfrac{1}{a²-(3+2)a+6} -\dfrac{2}{a²-(3+1)a+3} -\dfrac{1}{a²-(2+1)a+2}$
$= \dfrac{1}{a²-3a-2a+6} -\dfrac{2}{a²-3a-a+3} -\dfrac{1}{a²-2a-a+2}$
$= \dfrac{1}{a(a-3) -2(a-3)} -\dfrac{2}{a(a-3) -(a-3)} -\dfrac{1}{a(a-2) -1(a-2)}$
$= \dfrac{1}{(a-3)(a-2)} -\dfrac{2}{(a-3)(a-1)} -\dfrac{1}{(a-2)(a-1)}$
$= \dfrac{1(a-1) -2(a-2) -1(a-3)}{(a-1)(a-2)(a-3)}$
$= \dfrac{a-1 -2a+4 -a+3}{(a-1)(a-2)(a-3)}$
$= \dfrac{-2a+6}{(a-1)(a-2)(a-3)}$
$= \dfrac{-2(a-3)}{(a-1)(a-2)(a-3)}$
$= \dfrac{-2}{(a-1)(a-2)}$
$= \dfrac{2}{(1-a)(a-2)}$
= Answer
In a rational expressions, we can rewrite a negative expression as a positive expression by inverting one of the given term.
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