Question

Sabendo que as expressões 1+ax/1-ax e 5+a^2x^2/1-a^2x^2 são iguais, escreva o valor do número real x.

Answer 1

$\frac{1+ax}{1-ax} = \frac{5+ a^{2} x^{2} }{ 1- a^{2} x^{2} }$
$(1+ax)[1-(ax)^{2}] = [5+(ax)^2](1-ax)[tex] [tex]1-(ax)^{2}+ax-(ax)^{3} = 5 - 5ax+(ax)^{2} -ax^{3}$^{2}](1-ax) [/tex]
$-2a^2x^2 + 6ax - 4 = 0$
$\frac{-6a [tex] \frac{-6+2a}{-4a^{2}}=x \\ \frac{x = -3 +- a}{2a^{2}}$± \sqrt{36a^{2}-32a^{2}} }{-4a^{2}} = x[/tex]