Rozwiązać nierównosci
: 18 lis 2009, o 21:34
\(\displaystyle{ x^2-1>0 \wedge x>0 \wedge -1 \le -1+2log_8x \le 1}\)
\(\displaystyle{ x \neq 0 \wedge \frac{x-1}{x} \ge 0 \wedge x-1>0 \wedge -1 \le 2 log_{0,5}(x-1) \le 1}\)
\(\displaystyle{ x \neq 0 \wedge \frac{x-1}{x} \ge 0 \wedge x-1>0 \wedge -1 \le 2 log_{0,5}(x-1) \le 1}\)