Question

Gente eu queria saber como se faz essa conta:
$\frac{1-4x}{x} = \frac{7 -20x}{1 +4x}+ \frac{3}{x+4 x^{2} }$

Answer 1

$\dfrac{1-4x}{x}=\dfrac{7-20x}{1+4x}+\dfrac{3}{x+4x^{2}}\\\\\\ \dfrac{1-4x}{x}=\dfrac{7-20x}{1+4x}+\dfrac{3}{x(1+4x)}\\\\\\ \dfrac{(1-4x)(1+4x)}{x(1+4x)}=\dfrac{(7-20x)x}{(1+4x)x}+\dfrac{3}{x(1+4x)}\\\\ (1-4x)(1+4x)=(7-20x)x+3\\\\ 1-16x^{2}=7x-20x^{2}+3\\\\ 4x^{2}-7x-2=0$

$\Delta=b^{2}-4\cdot a\cdot c\\ \Delta=(-7)^{2}-4\cdot4\cdot(-2)\\ \Delta=49+32\\ \Delta=81\\\\ x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\\x=\dfrac{7\pm\sqrt{81}}{2\cdot4}\\\\x=\dfrac{7\pm9}{8}$

$\Longrightarrow x_{1}=\dfrac{7+9}{8}=\dfrac{16}{8}=2\\\\ \Longrightarrow x_{2}=\dfrac{7-9}{8}=\dfrac{-2}{8}=-\dfrac{1}{4}$

$S=\{-\dfrac{1}{4},\;2\}$