Question

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

Answer 1

$left { {frac{x-2}{4}+frac{x+4}{8}leq 6} atop {(x-4)^{2}<(x-1)(x-3)-5}} right.\\left { {{frac{x-2}{4}*8+frac{x+4}{8} *8leq 6*8  } atop {x^{2}-8x+16<x^{2}-3x-x+3-5  }} right.\\left { {{2(x-2)+x+4leq48 } atop {x^{2}-8x-x^{2}+4x<-2-16  }} right.\\left { {{2x-4+x+4leq48 } atop {-4x<-18}} right.\\left { {{3xleq 48} atop {x>4,5}} right. \\left { {{xleq16 } atop {x>4,5}} right.$

Ответ : x ∈ (4,5 ; 16]