Question

$1 + \frac{2}{x-4}= \frac{8}{ x^{2} -4x}$

Answer 1

Olá Jhennifer!!

$\\ \displaystyle \mathsf{1+\frac{2}{x-4}=\frac{8}{x^2-4x}}\\\\\\ \mathsf{1 + \frac{2}{x-4}=\frac{8}{x(x-4)}} \\\\\\ \mathsf{\frac{1}{1} + \frac{2}{(x - 4)} = \frac{8}{x(x - 4)} \qquad \qquad \qquad MMC(1, x - 4, x^2 - 4x) = x^2 - 4x} \\\\\\ \mathsf{\frac{1}{1/x(x - 4)} + \frac{2}{(x - 4)/x} = \frac{8}{x(x - 4)/1}} \\\\\\ \mathsf{x(x - 4) + 2 \cdot x = 8 \cdot 1} \\\\ \mathsf{x^2 - 4x + 2x = 8} \\\\ \mathsf{x^2 - 2x - 8 = 0} \\\\ \mathsf{x^2 - 4x + 2x - 8 = 0} \\\\ \mathsf{x(x - 4) + 2(x - 4) = 0} \\\\ \mathsf{(x - 4) \cdot (x + 2) = 0} \\\\ \boxed{\mathsf{S = \left \{ - 2, 4 \right \}}}$