решить неравенство алгебраическим и графическим способами

$||x-4|-2|<3$
$\left[\begin{array}{ccc} \left \{ {{x-4 \geq 0} \atop {|x-4-2|<3}} \right.\\ \left \{ {{x-4<0} \atop {|-x+4-2|<3}} \right. \end{array}\right$
Решаем отдельно
$\left \{ {{x-4 \geq 0} \atop { \left[\begin{array}{ccc} \left \{ {{x-4-2 \geq 0} \atop {x-4-2<3}} \right. \\ \left \{ {{x-4-2<0} \atop {-x+4+2<3}} \right. \end{array}\right}} \right.$
$\left \{ {{x-4-2 \geq 0} \atop {x-4-2<3}} \right. \to \left \{ {{x \geq 6} \atop {x<9}} \right. \to[6;9)$
Вторая система

3}} \right. \to (3;6)" alt=" \left \{ {{x-4-2<0} \atop {-x+4+2<3}} \right. \to \left \{ {{x<6} \atop {x>3}} \right. \to (3;6)" align="absmiddle" class="latex-formula">
Теперь
$\left \{ {{x-4<0} \atop { \left[\begin{array}{ccc} \left \{ {{-x+4-2 \geq 0} \atop {-x+4-2<3}} \right. \\ \left \{ {{-x+4-2<0} \atop {x-4+2<3}} \right. \end{array}\right}} \right.$
Решив отдельно, получаем

-1}} \right. \to(-1;2]" alt=" \left \{ {{-x+4-2 \geq 0} \atop {-x+4-2<3}} \right. \to \left \{ {{x \leq 2} \atop {x>-1}} \right. \to(-1;2]" align="absmiddle" class="latex-formula">
Второе

2} \atop {x<5}} \right. \to (-1;5)" alt=" \left \{ {{-x+4-2<0} \atop {x-4+2<3}} \right. \to \left \{ {{x>2} \atop {x<5}} \right. \to (-1;5)" align="absmiddle" class="latex-formula">
Объеденим и получаем ответ

Ответ: $(-1;9)$