Question

3-xПод корнем >x-1
Помогите пожалуйста

Answer 1

$\sqrt{3-x}>x-1\; \; \; \Leftrightarrow \; \; \; \left \{ {{x-1\geq 0} \atop {3-x>(x-1)^2}} \right.\; \; \; ili\; \; \; \left \{ {{x-1<0} \atop {3-x\geq 0}} \right.\\\\\left \{ {{x\geq 1\qquad \qquad } \atop {3-x>x^2-2x+1}} \right.\; \; \; ili\; \; \; \left \{ {{x<1} \atop {x\leq 3}} \right.\\\\\left \{ {{x\geq 1\qquad \; } \atop {x^2-x-2<0}} \right.\; \; \; \; \; ili\qquad x<1\\\\\left \{ {{x\geq 1} \atop {(x-2)(x+1)<0}} \right. \qquad ili\qquad x<1\\\\\left \{ {{x\geq 1} \atop {-1<x<2}} \right. \qquad ili\qquad x<1$

$x\in [\, 1,2)\qquad ili\qquad x\in (-\infty ,1)\\\\Otvet:\; \; x\in (-\infty ,1)\cup [\, 1,2)=(-\infty ,2)\; .$