Question

como resuelvo esta ecuacion con fracciones 5/(x+1)^2+4/(x-1)=1

Answer 1

$\frac{5}{\left(x+1\right)^2}+\frac{4}{\left(x-1\right)}=1$
Encontrar el mínimo común múltiplo de: $(x+1\right)^2,\:x-1:\quad \left(x+1\right)^2\left(x-1\right)$
$\mathrm{Multiplicar\:por\:el\:minimo\:comun\:multiplo=}\left(x+1\right)^2\left(x-1\right)$
$\frac{5}{\left(x+1\right)^2}\left(x+1\right)^2\left(x-1\right)+\frac{4}{x-1}\left(x+1\right)^2\left(x-1\right)=1\cdot \left(x+1\right)^2\left(x-1\right)$
$5\left(x-1\right)+4\left(x+1\right)^2=\left(x+1\right)^2\left(x-1\right)$
$Desarrollando\:tenemos: 3x^2+14x-x^3=0$
$\mathrm{Factorizar\:}3x^2+14x-x^3:\quad -x\left(x^2-3x-14\right)$
$-x\left(x^2-3x-14\right)=0$
$x_{1} =0$
$\:x^2-3x-14=0:\quad x_{2} =\frac{3+\sqrt{65}}{2},\: x_{3} =\frac{3-\sqrt{65}}{2}$