Question

Решить неравенство log x^2 (1/x + 2/x^2)<0

Answer 1

$log_{x^2} (frac{1}{x}+frac{2}{x^2}) textless 0, \ left[begin{array}{cc} left{begin{array}{ccc}frac{1}{x}+frac{2}{x^2} textgreater 0,\0 textless x^2 textless 1,\frac{1}{x}+frac{2}{x^2} textgreater 1;end{array}right. \ left{begin{array}{ccc}frac{1}{x}+frac{2}{x^2} textgreater 0,\x^2 textgreater 1,\frac{1}{x}+frac{2}{x^2} textless 1;end{array}right. end{array}right.$
$left[begin{array}{cc} left{begin{array}{cc} left [ {{-1 textless x textless 0,} atop {0 textless x textless 1;}} right.\frac{1}{x}+frac{2}{x^2}-1 textgreater 0;end{array}right. \ left{begin{array}{ccc}frac{x+2}{x^2} textgreater 0,\ left [ {{x textless -1,} atop {x textgreater 1;}} right. \frac{1}{x}+frac{2}{x^2}-1 textless 0;end{array}right. end{array}right.$
$left[begin{array}{cc} left{begin{array}{cc} left [ {{-1 textless x textless 0,} atop {0 textless x textless 1;}} right.\-x^2+x+2 textgreater 0;end{array}right. \ left{begin{array}{ccc} left { {{x+2 textgreater 0,} atop {xneq0;}} right. \ left [ {{x textless -1,} atop {x textgreater 1;}} right. \-x^2+x+2 textless 0;end{array}right. end{array}right.$
$left[begin{array}{cc} left{begin{array}{cc} left [ {{-1 textless x textless 0,} atop {0 textless x textless 1;}} right.\(x+1)(x-2) textless 0;end{array}right. \ left{begin{array}{cc} left [ {{-2 textless x textless -1,} atop {x textgreater 1;}} right. \(x+1)(x-2) textgreater 0;end{array}right. end{array}right.$
$left[begin{array}{cc} left{begin{array}{cc} left [ {{-1 textless x textless 0,} atop {0 textless x textless 1;}} right.\-1 textless x textless 2;end{array}right. \ left{begin{array}{cc} left [ {{-2 textless x textless -1,} atop {x textgreater 1;}} right. \ left [{ {{x textless -1,} atop {x textgreater 2;}} right. end{array}right. end{array}right.$
$left[begin{array}{cccc} -1 textless x textless 0,\0 textless x textless 1,\-2 textless x textless -1,\x textgreater 2;end{array}right. \ xin(-2;-1)cup(-1;0)cup(0;1)cup(2;+infty).$