Question

$log_{4-x}(16- x^{2} ) \leq 1$

Answer 1

Метод рационализации.

$log_{4-x}(16-x^2) \leq 1\; ,\; \; \; ODZ:\; \left \{ {{4-x\ \textgreater \ 0,\; 4-x\ne 1} \atop {16-x^2\ \textgreater \ 0}} \right. \; \left \{ {{x\ \textless \ 4,\; x\ne 3} \atop {x\in (-4,4)}} \right. \; \to \\\\x\in (-4,3)U(3,4)\\\\(4-x-1)(16-x^2-(4-x)) \leq 0\\\\(3-x)(-x^2+x+12) \leq 0\\\\(x-3)(x-4)(x+3) \leq 0\\\\---(-3)+++(3)---(4)+++\\\\x\in (-\infty,-3]U[3,4]\\\\ \left \{ {{x\in (-\infty,-3]U[3,4]} \atop {x\in (-4,3)U(3,4)}} \right. \; \; \to \; \; x\in (-4,3)U(3,4)$