Question

Answer 1

\(\dfrac{1+8x}{8x+4}=\dfrac{2x}{6x-3}-\dfrac{8x^2}{3-12x^2}\Leftrightarrow\dfrac{1+8x}{4\left(2x+1\right)}-\dfrac{2x}{3\left(2x-1\right)}+\dfrac{8x^2}{3\left(1-2x\right)\left(1+2x\right)}=0\Leftrightarrow\dfrac{3\left(1+8x\right)\left(1-2x\right)+8x\left(1+2x\right)+32x^2}{12\left(1-2x\right)\left(1+2x\right)}=0\Leftrightarrow3\left(1+8x\right)\left(1-2x\right)+8x\left(1+2x\right)+32x^2=0\Leftrightarrow-48x^2+18x+3+16x^2+8x+32x^2=0\Leftrightarrow26x+3=0\Leftrightarrow x=-\dfrac{3}{26}\)