Решить неравенство [tex] \frac{1-3^{ x^{2} +2x-3}}{x^{2} +2x-3} \leq 0[/tex]

Решение смотри на фото.

[tex] \frac{1-3^{x^2+2x-3}}{x^2+2x-3} \leq 0 [/tex]
let's [tex]t=x^2+2x-3[/tex]

[tex] \frac{1-3^t}{t} \leq 0; [/tex]

We have three possible cases:

first:

[tex] \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[/tex]

[tex]t=x^2+2x-3\ \textgreater \ 0[/tex]

[tex]x^2+3x-x-3\ \textgreater \ 0[/tex]

[tex]x(x+3)-(x+3)\ \textgreater \ 0[/tex]

[tex](x-1)(x+3)\ \textgreater \ 0[/tex]

[tex][x-(1)]*[x-(-3)]\ \textgreater \ 0[/tex]

[tex]x\in (-\infty;-3)\cup(1;+\infty)[/tex]

second:

[tex] \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[/tex]

[tex]t=x^2+2x-3 \ \textless \ 0[/tex]

[tex][x-(1)]*[x-(-3)]\ \textless \ 0[/tex]

[tex]x\in (-3;1)[/tex]

third:

[tex] \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. [/tex]

The system of inequalities behind have not sense due to its first inequality.
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So, we have: [tex]t\in (-\infty;0)\cup(0;+\infty)[/tex]

and [tex]x^2+2x-3 \neq 0;[/tex]

[tex](x-1)(x+3) \neq 0[/tex]

[tex]x\in (-\infty;-3)\cup(-3;1)\cup(1;+\infty)[/tex]

Answer: [tex](-\infty;-3)\cup(-3;1)\cup(1;+\infty)[/tex]