Question

(x^2-9)*log x по основанию 1/3 < или = 0

Answer 1

$(x^2-9)log_{\frac{1}{3}}x \leq 0\; ,\; OOF:\; x0\\\\a)\; \left \{ {{x^2-9 \leq 0} \atop {log_{\frac{1}{3}}x \geq 0}} \right. \; \left \{ {{(x-3)(x+3) \leq 0} \atop {log_{\frac{1}{3}}x \geq log_{\frac{1}{3}}1}} \right. \left \{ {{x\in [-3,3]} \atop {x\in (-\infty,1],x0}} \right. \; \to \; x\in(0,1]\\\\b)\; \left \{ {{x^2-9 \geq 0} \atop {log_{\frac{1}{3}}x \leq 0}} \right. \; \left \{ {{x\in (-\infty,-3]U[3,+\infty)} \atop {x\in [1,+\infty),x0}} \right. \; \; \to \; \; x\in [3,+\infty)$

$Otvet:\; x\in (0,1]U[3,+\infty)$