Question

Решите систему неравенств: x-2/4+x+4/8<=6 (x-4)^2<(x-1)(x-3)-5

Answer 1

$\left \{ {\frac{x-2}{4}+\frac{x+4}{8}\leq 6} \atop {(x-4)^{2}<(x-1)(x-3)-5}} \right.\\\\\left \{ {{\frac{x-2}{4}*8+\frac{x+4}{8} *8\leq 6*8  } \atop {x^{2}-8x+16<x^{2}-3x-x+3-5  }} \right.\\\\\left \{ {{2(x-2)+x+4\leq48 } \atop {x^{2}-8x-x^{2}+4x<-2-16  }} \right.\\\\\left \{ {{2x-4+x+4\leq48 } \atop {-4x<-18}} \right.\\\\\left \{ {{3x\leq 48} \atop {x>4,5}} \right. \\\\\left \{ {{x\leq16 } \atop {x>4,5}} \right.$

Ответ : x ∈ (4,5 ; 16]