Question

2/3>(4/9)^3x^-1 atrisināt nevienādību

Answer 1

$\frac{2}{3}>\Big (\frac{4}{9}\Big )^{3x^{-1}}\\\\\frac{2}{3}>\Big (\frac{2}{3}\Big )^{6x^{-1}}\\\\Tak\; kak\; \; a=\frac{2}{3}<1\; ,\; to\; \; \; 1<6x^{-1}\; \; \Rightarrow \; \; \frac{6}{x}>1\; \; ,\; \; \frac{6}{x}-1>0\; \; ,\; \frac{6-x}{x}>0\; ,\\\\\frac{x-6}{x}<0\; \qquad znaki:\; \; +++(0)---(6)+++\\\\\underline {\; x\in (0,6)\; }$