Question

giải pt sau:

\(\frac{3}{1-4x}=\frac{2}{4x+1}-\frac{8+6x}{16x^2-1}\)

Answer 1

\(\frac{3}{1-4x}\) = \(\frac{2}{4x+1}\) - \(\frac{8+6x}{16x^2-1}\)

<=>\(\frac{-3\left(4x+1\right)}{\left(4x+1\right)\left(4x-1\right)}\)=\(\frac{2\left(4x-1\right)}{\left(4x+1\right)\left(4x-1\right)}\)- \(\frac{8+6x}{16x^2-1}\)

<=>\(\frac{-12x-3}{16x^2-1}\)=\(\frac{8x-2}{16x^2-1}\)- \(\frac{8+6x}{16x^2-1}\)

<=> -12x - 3 = 8x-2-8-6x

<=> -12x - 3 = 2x-10

<=> 3-10 = -12x-2x

<=> -7 = -14x

<=> x = 0,5