Question

Решите неравенство
|x^3-1|>1-x

Answer 1

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