Question

Frações Algébricas - 3* caso D" id="TexFormula1" title="\frac{x+1}{2x-4} + \frac{x-1}{3x-6}" alt="\frac{x+1}{2x-4} + \frac{x-1}{3x-6}" align="absmiddle" class="latex-formula">

Answer 1

Answer:

$\dfrac{5x + 1}{6(x - 2)}$

$\dfrac{5x + 1}{6x - 12}$

Step-by-step explanation:

Fatorizam-se os denominadores.

2x - 4 = 2(x - 2)

3x - 6 = 3(x - 2)

O minimo multiplo comum dos denominadores e

2(3)(x - 2)

$\dfrac{x + 1}{2x - 4} + \dfrac{x - 1}{3x - 6} =$

$= \dfrac{3(x + 1)}{2(3)(x - 2)} + \dfrac{2(x - 1)}{2(3)(x - 2)}$

$= \dfrac{3(x + 1) + 2(x - 1)}{6(x - 2)}$

$= \dfrac{3x + 3 + 2x - 2)}{6(x - 2)}$

$= \dfrac{5x + 1}{6(x - 2)}$

$= \dfrac{5x + 1}{6x - 12}$