Question

simplificati fractia: 3x²-4 pe x³-8 impartit la √3x+2 pe x²-4

Answer 1

$\frac{3x ^{2}-4 }{x ^{3} -8} : \frac{ \sqrt{3} x+2}{x ^{2} -4} = \frac{ (\sqrt{3} x-2)( \sqrt{3}x+2) }{(x-2)(x ^{2}+2x+4) } * \frac{(x-2)(x+2)}{ \sqrt{3}x+2 }$
$= \frac{ \sqrt{3} x-2}{ x^{2} +2x+4} * (x+2)= \frac{ (\sqrt{3} x-2)(x+2)}{ x^{2} +2x+4}$