Question

Resuelva la siguiente ecuación:

$\frac{4x+1}{4x-1} - \frac{6}{16 x^{2}-1 } = \frac{4x-1}{4x+1}$

Answer 1

$\frac{4x+1}{4x-1}-\frac{6}{16x^2-1}=\frac{4x-1}{4x+1}$
$\mathrm{Encontrar\:el\:minimo\:comun\:multiplo\:de\:}4x-1,\:16x^2-1,\:4x+1:\quad \left(4x-1\right)\left(4x+1\right)$
$\mathrm{Multiplicar\:por\:el\:minimo\:comun\:multiplo=}\left(4x-1\right)\left(4x+1\right)$
$\frac{4x+1}{4x-1}\left(4x-1\right)\left(4x+1\right)-\frac{6}{16x^2-1}\left(4x-1\right)\left(4x+1\right)=\frac{4x-1}{4x+1}\left(4x-1\right)\left(4x+1\right)$
$\left(4x+1\right)^2-6=\left(4x-1\right)^2$
$\mathrm{Aplicar\:la\:siguiente\:regla\:de\:productos\:notables}:\quad \left(a+b\right)^2=a^2+2ab+b^2$
$\mathrm{Aplicar\:la\:siguiente\:regla\:de\:productos\:notables}:\quad \left(a-b\right)^2=a^2-2ab+b^2$
$16x^2+8x-5-6=16x^2-8x+1$
$16x^2+8x-5=16x^2-8x+1$
$\mathrm{Sumar\:}5\mathrm{\:a\:ambos\:lados}:$
$16x^2+8x-5+5=16x^2-8x+1+5$
$16x^2+8x=16x^2-8x+6$
$\mathrm{Restar\:}16x^2-8x\mathrm{\:de\:ambos\:lados}:$
$16x^2+8x-\left(16x^2-8x\right)=16x^2-8x+6-\left(16x^2-8x\right)$
$16x=6$
$\mathrm{Dividir\:ambos\:lados\:entre\:}16:$
$\frac{16x}{16}=\frac{6}{16}$
$x=\frac{3}{8}\quad \left(\mathrm{Decimal:\quad }x=0.375\right)$