Rešenje:
$\sqrt{x^2+3x+2} - \sqrt{x^2-x+1} < 1$
Uslovi:
$\begin{array}{|l} x^2+3x+2 \ge 0 \\ x^2-x+1 \ge 0 \end{array}$
$x^2+3x+2 \ge 0$
$x^2+3x+2 = 0$
$D=1$
$x_{1}=-1 ; x_{2}=-2$
$x^2+3x+2 \ge 0 \Leftrightarrow x \in (-\infty,-2) \cup (-1,+\infty)$
$x^2-x+1 \ge 0$
$x^2-x+1 = 0$
$D=-3 \Rightarrow$
$x^2-x+1 \ge 0$ for $\forall x$
$\begin{array}{|l} x^2+3x+2 \ge 0 \\ x^2-x+1 \ge 0 \end{array} \Leftrightarrow
\begin{array}{|l} \forall x \\ x \in (-\infty,-1) \cup (-2,+\infty) \end{array} \Leftrightarrow
x \in (-\infty,-2) \cup (-1,+\infty)$
$\sqrt{x^2+3x+2} - \sqrt{x^2-x+1} < 1$
$\sqrt{x^2+3x+2} < 1 + \sqrt{x^2-x+1} \uparrow^2$
$x^2+3x+2 < 1+2\sqrt{x^2-x+1}+x^2-x+1$
$4x<2\sqrt{x^2-x+1} /:2$
$2x<\sqrt{x^2-x+1}$
$\sqrt{x^2-x+1}>2x$
$\Leftrightarrow$
$\left [
\begin{array}{l}
\begin{array}{|l} x^2-x+1 \ge 0 \\ 2x < 0 \end{array} \\
\begin{array}{} \end{array} \\
\begin{array}{|l} x^2-x+1 \ge 0 \\ 2x \ge 0 \\ x^2-x+1 \ge 4x^2 \end{array}
\end{array}
\right. \Leftrightarrow$
$\left [
\begin{array}{l}
\begin{array}{|l} \forall x \\ x < 0 \end{array} \\
\begin{array}{} \end{array} \\
\begin{array}{|l} \forall x \\ x \ge 0 \\ 3x^2+x-1 < 0 \end{array}
\end{array}
\right. \Leftrightarrow$
$3x^2+x-1 < 0$
$3x^2+x-1 = 0$
$D=13$
$x_{1/2}=\frac{-1 \pm \sqrt{13}}{6} \approx$
$x_{1}=\frac{-1 + \sqrt{13}}{6} \approx 0,43$
$x_{2}=\frac{-1 - \sqrt{13}}{6} \approx 0,76$
$3x^2+x-1 < 0 \Leftrightarrow x \in (\frac{-1 - \sqrt{13}}{6},\frac{-1 + \sqrt{13}}{6})$
$\Rightarrow$
$\left [
\begin{array}{l}
\begin{array}{|l} \forall x \\ x < 0 \end{array} \\
\begin{array}{} \end{array} \\
\begin{array}{|l} \forall x \\ x \ge 0 \\ 3x^2+x-1 < 0 \end{array}
\end{array}
\right. \Leftrightarrow$
$\left [
\begin{array}{l}
\begin{array}{l} x \in (-\infty,0) \end{array} \\
\begin{array}{} \end{array} \\
\begin{array}{|l} x \ge 0 \\ x \in (\frac{-1 - \sqrt{13}}{6},\frac{-1 + \sqrt{13}}{6}) \end{array}
\end{array}
\right. \Leftrightarrow$
$\left [
\begin{array}{l}
\begin{array}{l} x \in (-\infty,0) \end{array} \\
\begin{array}{} \end{array} \\
\begin{array}{l} x \in (0,\frac{-1 + \sqrt{13}}{6}) \end{array}
\end{array}
\right. \Leftrightarrow$
$x \in (-\infty,\frac{-1 + \sqrt{13}}{6})$
No
uslov: $x\in (-\infty,-2) \cup (-1,+\infty) \Rightarrow$
$x \in (-\infty,-2) \cup (-1,\frac{-1 + \sqrt{13}}{6})$
Odgovor: $x \in (-\infty,-2) \cup (-1,\frac{-1 + \sqrt{13}}{6})$