0}} \right. \left \{ {{x \leq -20} \atop {x>-7}} \right. \\ \\ \left \{ {{x+20 \geq0 } \atop {7+x<0}} \right. \left \{ {{x \geq -20} \atop {x<-7}} \right. " alt=" \frac{x-6}{-7-x} \leq -2 \\ \\ \frac{x-6+2(-7-x)}{-7-x} \leq 0 \\ \\ \frac{x-6-14-2x}{-7-x} \leq 0 \\ \\ \frac{-x-20}{-7-x} \leq 0 \\ \\ \frac{x+20}{7+x} \leq 0 \\ \\ \\ \left \{ {{x+20 \leq 0} \atop {7+x>0}} \right. \left \{ {{x \leq -20} \atop {x>-7}} \right. \\ \\ \left \{ {{x+20 \geq0 } \atop {7+x<0}} \right. \left \{ {{x \geq -20} \atop {x<-7}} \right. " align="absmiddle" class="latex-formula">
первая система общих корней не имеет
x∈[-20; -7)
13 целых корней
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1 }} \right. " alt=" \left \{ {{|x-1| \leq 6} \atop {( \frac{1}{7})^{0,5x-1}>1 }} \right. " align="absmiddle" class="latex-formula">
1. Пусть |x-1| при раскрытии примем <0 => x-1<0 => x<1, тогда<br>
1 }} \right. \left \{ {{x \geq -5} \atop {( \frac{1}{7})^{0,5x-1}>( \frac{1}{7})^0 }} \right. \left \{ {{x \geq -5} \atop { 0,5x-1<0 }} \right. \left \{ {{x \geq -5} \atop {x < 2}} \right. " alt="\left \{ {{-x+1 \leq 6} \atop {( \frac{1}{7})^{0,5x-1}>1 }} \right. \left \{ {{x \geq -5} \atop {( \frac{1}{7})^{0,5x-1}>( \frac{1}{7})^0 }} \right. \left \{ {{x \geq -5} \atop { 0,5x-1<0 }} \right. \left \{ {{x \geq -5} \atop {x < 2}} \right. " align="absmiddle" class="latex-formula">
6 целых корней (1 не входит, т.к. мы считаем модуль отрицательным)
2. Пусть
x \geq 1" alt="|x-1| \geq 0 => x \geq 1" align="absmiddle" class="latex-formula">
( \frac{1}{7} )^0 }} \right.\left \{ {{x \leq 7} \atop { 0,5x-1<0}} \right. \left \{ {{x \leq 7} \atop {x<2}} \right. " alt="\left \{ {{x-1 \leq 6} \atop {( \frac{1}{7})^{0,5x-1}>( \frac{1}{7} )^0 }} \right.\left \{ {{x \leq 7} \atop { 0,5x-1<0}} \right. \left \{ {{x \leq 7} \atop {x<2}} \right. " align="absmiddle" class="latex-formula">
Количество целых корней от 1 до 2(не включительно) = 1 корень
Ответ: 7 корней