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

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  • \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}} ight. ; \left \{ {{1-3^t \leq 0} \atop {t\ \textgreater \ 0}} ight. ; \left \{ {{-3^t \leq -1} \atop {t\ \textgreater \ 0}} ight. ; \left \{ {{3^t \geq 1} \atop {t\ \textgreater \ 0}} ight. ; \left \{ {{3^t \geq 3^0} \atop {t\ \textgreater \ 0}} ight. ; \left \{ {{t \geq 0} \atop {t\ \textgreater \ 0}} ight. ; t\ \textgreater \ 0t=x^2+2x-3\ \textgreater \ 0x^2+3x-x-3\ \textgreater \ 0x(x+3)-(x+3)\ \textgreater \ 0(x-1)(x+3)\ \textgreater \ 0[x-(1)]*[x-(-3)]\ \textgreater \ 0x\in (-\infty;-3)\cup(1;+\infty)second: \left \{ {{\frac{1-3^t}{t} \leq 0|*t} \atop {t\ \textless \ 0}} ight. ; \left \{ {{1-3^t \geq 0} \atop {t\ \textless \ 0}} ight. ; \left \{ {{-3^t \geq -1} \atop {t\ \textless \ 0}} ight. ; \left \{ {{3^t \leq 1} \atop {t\ \textless \ 0}} ight. ; \left \{ {{3^t \leq 3^0} \atop {t\ \textless \ 0}} ight. ; \left \{ {{t \leq 0} \atop {t\ \textless \ 0}} ight. ; t\ \textless \ 0t=x^2+2x-3 \ \textless \ 0[x-(1)]*[x-(-3)]\ \textless \ 0x\in (-3;1)third: \left \{ {{ \frac{1-3^t}{t} \leq 0} \atop {t=0}} ight. ; \left \{ {{ \frac{1-3^0}{0} \leq 0} \atop {t=0}} ight. ; \left \{ {{ \frac{0}{0} \leq 0} \atop {t=0}} ight. The system of inequalities behind have not sense due to its first inequality.----------------------------------So, we have: t\in (-\infty;0)\cup(0;+\infty)and x^2+2x-3 eq 0;(x-1)(x+3) eq 0x\in (-\infty;-3)\cup(-3;1)\cup(1;+\infty)Answer: (-\infty;-3)\cup(-3;1)\cup(1;+\infty)

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