4,5}} \right. \\\\\left \{ {{x\leq16 } \atop {x>4,5}} \right." alt="\left \{ {\frac{x-2}{4}+\frac{x+4}{8}\leq 6} \atop {(x-4)^{2}<(x-1)(x-3)-5}} \right.\\\\\left \{ {{\frac{x-2}{4}*8+\frac{x+4}{8} *8\leq 6*8 } \atop {x^{2}-8x+16<x^{2}-3x-x+3-5 }} \right.\\\\\left \{ {{2(x-2)+x+4\leq48 } \atop {x^{2}-8x-x^{2}+4x<-2-16 }} \right.\\\\\left \{ {{2x-4+x+4\leq48 } \atop {-4x<-18}} \right.\\\\\left \{ {{3x\leq 48} \atop {x>4,5}} \right. \\\\\left \{ {{x\leq16 } \atop {x>4,5}} \right." align="absmiddle" class="latex-formula">
Ответ : x ∈ (4,5 ; 16]