Question

решить неравенство .

Answer 1

$\frac{x^2+3x-4}{4^{x}+2^{x+1}-8} \geq 0\\\\a)\; \; x^2+3x-4=0\; \; \to \; \; x_1=-4,\; x_2=1\\\\x^2+3x-4=(x+4)(x-1)\\\\b)\; \; 4^{x}+2^{x+1}-8=0\; ,\; (2^{x})^2+2^{x}\cdot 2-8=0\; ,\\\\t=2^{x}\; \; \to \; \; t^2+2t-8=0\; \to \; t_1=-4\; ,\; t_2=2\; \; \to \\\\ t^2+2t-8=(t+4)(t-2)\\\\4^{x}+2^{x+1}-8=(2^{x}+4)(2^{x}-2)\\\\2^{x}\ \textgreater \ 0\; \; \Rightarrow \; \; 2^{x}+4\ \textgreater \ 0\\\\2^{x}-2\ \textgreater \ 0\; ,2^{x}\ \textgreater \ 2^1\; ,\; x\ \textgreater \ 1\\\\2^{x}-2\ \textless \ 0\; \; pri\; \; x\ \textless \ 1$

$c)\; \; \frac{(x+4)(x-1)}{(2^{x}+4)(2^{x}-2)} \geq 0$

$\left \{ {{(x+4)(x-1) \geq 0} \atop {(2^{x}+4)(2^{x}-2)\ \textgreater \ 0}} \right. \; \; ili\; \; \left \{ {{(x+4)(x-1) \leq 0} \atop {(2^{x}+4)(2^{x}-2)\ \textless \ 0}} \right.$

$Znaki\; chislitelya:\; \; +++[-4]---[1]+++$

$Znaki\; znam.:\; ---(1)+++$

$\left \{ {{x\in (-\infty ,-4]U[\, 1,+\infty )} \atop {x\in (1,+\infty )}} \right. \; \; ili\; \; \left \{ {{x\in [-4,1\, ]} \atop {x\in (-\infty ,1)}} \right. \\\\x\in (1,+\infty )\; \; \; ili \; \; \; x\in [-4,1)\\\\Otvet:\; \; x\in [-4,1)U(1,+\infty)$