Respostas

2013-10-24T19:12:10-02:00
Hola.

9 \leq  \frac{16}{(x-1)*(x+1)}\\ 
9*(x-1)*(x+1) \leq 16\\
9*(x^2 -1) \leq 16\\
(x^2-1) \leq  \frac{16}{9}\\ t{ (x^2-1) \leq\frac{16}{9}}\\
x^2 - 1  \leq  \frac{16}{9} \\
x^2 \leq  \frac{16}{9} +1 \\
x^2 \leq \frac{25}{9}
Extraindo a raiz quadrada em ambos os lados vc encontra:

x ≤ ± 5/3