Question

Решить неравенство $frac{1-3^{ x^{2} +2x-3}}{x^{2} +2x-3} leq 0$

Answer 1

$frac{1-3^{x^2+2x-3}}{x^2+2x-3} leq 0$
let's $t=x^2+2x-3$

$frac{1-3^t}{t} leq 0;$

We have three possible cases:

first:

$left { {{frac{1-3^t}{t} leq 0|*t} atop {t textgreater 0}} right. ; left { {{1-3^t leq 0} atop {t textgreater 0}} right. ; left { {{-3^t leq -1} atop {t textgreater 0}} right. ; left { {{3^t geq 1} atop {t textgreater 0}} right. ; left { {{3^t geq 3^0} atop {t textgreater 0}} right. ; left { {{t geq 0} atop {t textgreater 0}} right. ; t textgreater 0$

$t=x^2+2x-3 textgreater 0$

$x^2+3x-x-3 textgreater 0$

$x(x+3)-(x+3) textgreater 0$

$(x-1)(x+3) textgreater 0$

$[x-(1)]*[x-(-3)] textgreater 0$

$xin (-infty;-3)cup(1;+infty)$

second:

$left { {{frac{1-3^t}{t} leq 0|*t} atop {t textless 0}} right. ; left { {{1-3^t geq 0} atop {t textless 0}} right. ; left { {{-3^t geq -1} atop {t textless 0}} right. ; left { {{3^t leq 1} atop {t textless 0}} right. ; left { {{3^t leq 3^0} atop {t textless 0}} right. ; left { {{t leq 0} atop {t textless 0}} right. ; t textless 0$

$t=x^2+2x-3 textless 0$

$[x-(1)]*[x-(-3)] textless 0$

$xin (-3;1)$

third:

$left { {{ frac{1-3^t}{t} leq 0} atop {t=0}} right. ; left { {{ frac{1-3^0}{0} leq 0} atop {t=0}} right. ; left { {{ frac{0}{0} leq 0} atop {t=0}} right.$

The system of inequalities behind have not sense due to its first inequality.
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So, we have: $tin (-infty;0)cup(0;+infty)$

and $x^2+2x-3 neq 0;$

$(x-1)(x+3) neq 0$

$xin (-infty;-3)cup(-3;1)cup(1;+infty)$

Answer: $(-infty;-3)cup(-3;1)cup(1;+infty)$