Question

Решите неравенство
|x^3-1|>1-x

Answer 1

$|x^3-1| textgreater 1-x \ left[begin{array}{ccc}x^3-1 textgreater 1-x\x^3-1 textless -(1-x)end{array}right. Rightarrowleft[begin{array}{ccc}x^3+x-2 textgreater 0\x^3-1 textless x-1end{array}right. Rightarrow left[begin{array}{ccc}x^3+x-2 textgreater 0\x^3-x textless 0end{array}right. \x^3+x-2=0 \P(1)=1+1-2=0Rightarrow x_1=1 \x^3+x-2=(x-1)(x^2+ax+b)=x^3+ax^2+bx-x^2-ax-b=\=x^3+x^2(a-1)+x(b-a)-b \ a-1=0 \a=1 \-b=-2 \b=2 \x^3+x-2=(x-1)(x^2+x+2) \(x-1)(x^2+x+2) textgreater 0 \x^2+x+2 textgreater 0,forall x in R \x-1 textgreater 0 \x textgreater 1 \xin (1;+infty)$
$x^3-x textless 0 \x(x^2-1) textless 0 \x(x-1)(x+1) textless 0$
  -      +      -       +
----()-----()-----()--->
    -1      0      1
$xin (-infty;-1)cup (0;1) \ left[begin{array}{ccc}xin (-infty;-1)cup (0;1)\xin (1;+infty)end{array}right.Rightarrow x in(-infty;-1)cup(0;1)cup(1;+infty)$
Ответ: $x in(-infty;-1)cup(0;1)cup(1;+infty)$