Question

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

Answer 1

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

$Otvet:; xin (0,1]U[3,+infty)$