Question

найдите количество целых решений неравенства

Answer 1

$log_{2+x}(8-|x|) \geq 0\\\\ODZ:\; \; \left \{ {{8-|x|\ \textgreater \ 0} \atop {2+x\ \textgreater \ 0,\; 2+x\ne 1}} \right. \; \left \{ {{|x|\ \textless \ 8} \atop {x\ \textgreater \ -2,\; x\ne -1}} \right. \; \left \{ {{-8\ \textless \ x\ \textless \ 8} \atop {x\ \textgreater \ -2,\; x\ne -1}} \right. \; \Rightarrow \\\\\underline {x\in (-2,-1)\cup (-1,8)}$

Метод рационализации: 
заменяем  $log_{h}f\vee 0$  на  $(h-1)(f-1)\vee 0$  .

$(2+x-1)(8-|x|-1) \geq 0\\\\(x+1)(7-|x|) \geq 0\\\\a)\; \; x \geq 0:\; |x|=x\; \; \to \; \; (x+1)(7-x) \geq 0\; ,\\\\(x+1)(x-7) \leq 0\\\\+++[-1\, ]---[\, 7\, ]+++\; \; \; \; x\in [-1,7\, ]\\\\ \left \{ {{x \geq 0} \atop {x\in [-1,7\, ]}} \right. \; \; \to \; \; x\in [\, 0,7\, ]\; ,\; \; \left \{ {{x\on [\, 0,7\, ]} \atop {x\in (-2,-1)\cup (-1,8)}} \right. \; \; \to \; \; \underline {x\in [\, 0,7\, ]}\\\\b)\; \; x\ \textless \ 0:\; \; |x|=-x\; \; \to \; \; (x+1)(7+x) \geq 0\; ,\\\\+++(-7)---(-1)+++\; \; \; x\in (-\infty ,-7\, ]\cup [-1,+\infty )$

$\left \{ {{x\ \textless \ 0} \atop {x\in (-\infty ,-7\, ]\cup [-1,+\infty )}} \right. \; \; \to \; \; x\in (-\infty ,-7\, ]\cup [-1,0)\\\\ \left \{ {{x\in (-\infty ,-7\, ]\cup [-1,0)} \atop {x\in (-2,-1)\cup (-1,8)}} \right. \; \; \to \; \; \underline {x\in (-1,0)}\\\\c)\; \; \left \{ {{x\in [\, 0,7\, ]} \atop {x\in (-1,0)}} \right. \; \; \to \; \; \underline {\underline {x\in (-1,7\, ]\; }}\\\\Celue\; resheniya:0,1,2,3,4,5,6,7.\\\\Kolichestvo\; \; celux\; reshenij=8.$