Question

найдите количество целых решений неравенства

Answer 1

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

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

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

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