Question

Решите неравенство
х+2<√х+14

Answer 1

$x+2<\sqrt{x+14}\; \; \Rightarrow \\\\\sqrt{x+14}>x+2\; \; \Leftrightarrow \left \{ {{x+2\geq 0\qquad \quad } \atop {x+14>x^2+4x+4}} \right.\; \; ili\; \; \left \{ {{x+2<0} \atop {x+14\geq 0}} \right.\\\\\left \{ {{x\geq -2\qquad \; \; } \atop {x^2+3x-10<0}} \right.\; \; \; ili\; \; \; \left \{ {{x<-2} \atop {x\geq -14}} \right.\\\\\left \{ {{x\geq -2\qquad \; } \atop {(x-2)(x+5)<0}} \right.\; \; \; ili\; \; \; -14\leq x<-2\\\\\left \{ {{x\geq -2} \atop {-5<x<2}} \right.\; \; \; ili\; \; \; \; -14\leq x<-2$

$x\in [-2,2)\; \; \; \; \; ili\; \; \; \; \; x\in [-14,-2)\\\\Otvet:\; \; x\in [-14,-2)\cup [-2,2)=[-14,2)\; .$

Answer 2

Ответ: во вложении Объяснение: