Question

Simplifique:

$(\frac{x+1}{x-2}+ \frac{x-3}{x+2}): \frac{2 x^{2}-2x+4 }{x-2}$

Quero Resolução Se Possível

Agradeço desde já

Answer 1

Olá!

$\\ \mathsf{\left ( \frac{x + 1}{x - 2/_{x + 2}} + \frac{x - 3}{x + 2/_{x - 2}} \right ) \div \frac{2x^2 - 2x + 4}{x - 2} =} \\\\\\ \mathsf{\left [ \frac{(x + 2) \cdot (x + 1) + (x - 2) \cdot (x - 3)}{(x - 2)(x + 2)} \right ] \div \frac{2(x^2 - x + 2)}{x - 2} =} \\\\\\ \mathsf{\left [ \frac{x^2 + 3x + 2 + x^2 - 5x + 6}{(x - 2)(x + 2)} \right ] \cdot \frac{x - 2}{2(x^2 - x + 2)} =}$

$\\ \mathsf{\left [ \frac{2x^2 - 2x + 8}{(x - 2)(x + 2)} \right ] \cdot \frac{x - 2}{2(x^2 - x + 2)} =} \\\\\\ \mathsf{\frac{2 \cdot (x^2 - x + 4) \cdot (x - 2)}{2 \cdot (x - 2) \cdot (x + 2) \cdot (x^2 - x + 2)} =} \\\\\\ \boxed{\mathsf{\frac{x^2 - x + 4}{(x + 2)(x^2 - x + 2)}}}$