Question

Допоможіть, будь ласка.
:)

Answer 1

$\begin{cases}\left | x^2+5x \right | < 6\\ |x+1|\leq 1\end{cases}\Leftrightarrow \begin{cases}\left ( x^2+5x \right )^2 < 6^2\\ (x+1)^2\leq 1\end{cases}\Leftrightarrow \begin{cases}\left ( x^2+5x-6 \right )\left ( x^2+5x+6 \right ) < 0\\ (x+1-1)(x+1+1)\leq 0\end{cases}\Leftrightarrow$ $\Leftrightarrow \begin{cases}(x-1)(x+6)(x+3)(x+2) < 0\\ x(x+2)\leq 0\end{cases}\Rightarrow \begin{cases}x\in (-6,-3)\cup (-2,1)\\x\in [-2,0]\end{cases}\Rightarrow x\in (-2,0]$

$\begin{cases}\left | x^2-4x \right | < 5\\|x+1| < 3\end{cases}\Leftrightarrow \begin{cases}\left ( x^2-4x \right )^2 < 5^2\\(x+1)^2 < 3^2\end{cases}\Leftrightarrow \begin{cases}\left ( x^2-4x-5 \right )\left ( x^2-4x+5 \right ) < 0\\(x+1-3)(x+1+3) < 0\end{cases}\Leftrightarrow$

$\Leftrightarrow \begin{cases}(x+1)(x-5) < 0\\(x-2)(x+4) < 0\end{cases}\Rightarrow \begin{cases}x\in (-1,5)\\x\in (-4,2)\end{cases}\Rightarrow x\in (-1,2)$