Question

resolva as seguintes equações;

a)
${2x}^{2} - 8x + 6 = 0$
b)
$\frac{x}{x + 1} - \frac{x}{1 - x} = \frac{8}{3}$
​

Answer 1

Resposta:

$a)\\ \\x=\frac{-b+-\sqrt{b^{2}-4.a.c}}{2.a}\\ \\x=\frac{8+-\sqrt{(-8)^{2}-4.2.6}}{2.2}\\ \\x=\frac{8+-\sqrt{64-48}}{4}\\ \\x=\frac{8+-\sqrt{16}}{4}\\ \\x'=\frac{8+4}{4}=\frac{12}{4}=3\\ \\x''=\frac{8-4}{4}=\frac{4}{4}=1\\ \\Soulcao:~(x'=3;~~x''=1)$

$b)\\ \\\frac{x}{x+1}-\frac{x}{x-1}=\frac{8}{3}\\ \\mmc=(x+1).(x-1)\\ \\\frac{x^{2}-x-x^{2}+x}{(x+1).(x-1)}=\frac{8}{3}\\ \\\frac{0}{(x+1).(x-1)}=\frac{8}{3}\\ \\\frac{0}{x^{2}-1}=\frac{8}{3}\\ \\8.(x^{2}-1)=3.0\\ \\8x^{2}-8=0\\ \\8x^{2}=8\\ \\x^{2}=\frac{8}{8}\\ \\x^{2}=1\\ \\x=\sqrt{1}\\ \\x=1$