Absolute value inequality

Exercise:

Solve in $\mathbb{R},$ $\left| \frac{{{x}^{2}}-5x+4}{{{x}^{2}}-4} \right|\le 1$

Solution: we have \[\left| \frac{{{x}^{2}}-5x+4}{{{x}^{2}}-4} \right|\le 1\Leftrightarrow \left| {{x}^{2}}-5x+4 \right|\le \left| {{x}^{2}}-4 \right|\]

S.B.S  ${{\left( {{x}^{2}}-5x+4 \right)}^{2}}\le {{\left( {{x}^{2}}-4 \right)}^{2}}$ to get

$\Rightarrow {{x}^{4}}-10{{x}^{3}}+33{{x}^{2}}-40x+16\le {{x}^{4}}+16-8{{x}^{2}}$

$\Rightarrow -10{{x}^{3}}+33{{x}^{2}}-40x\le -8{{x}^{2}}$

$\Rightarrow -10{{x}^{3}}+41{{x}^{2}}-40x\le 0$

$\Rightarrow -x\left( 10{{x}^{2}}-41x+40 \right)\le 0\Leftrightarrow x\left( 10{{x}^{2}}-41x+40 \right)\ge 0$

But $10{{x}^{2}}-41x+40=\left( 5x-8 \right)\left( 2x-5 \right)$

Hence $x\left( 5x-8 \right)\left( 2x-5 \right)\ge 0$

Note that $x=0\,\,,\,\,\,x=\frac{8}{5}\,\,\,\,,\,\,\,x=\frac{5}{2}$ are the roots of $x\left( 5x-8 \right)\left( 2x-5 \right)$

Also remark that $f\left( x \right)=\frac{{{x}^{2}}-5x+4}{{{x}^{2}}-4}$ is defined when ${{x}^{2}}-4\ne 0\Leftrightarrow x\ne \pm 2$

So from the sign table we get

\[\begin{align}
  & \begin{matrix}
   x  \\
   x  \\
   5x-8  \\
   2x-5  \\
\end{matrix}\,\,\,\,\,\,\begin{matrix}
   -\infty   \\
   -  \\
   -  \\
   -  \\
\end{matrix}\,\,\,\,\,\,\,\,\begin{matrix}
   -2  \\
   ||  \\
   ||  \\
   ||  \\
\end{matrix}\,\,\,\,\,\,\,\begin{matrix}
   {}  \\
   -  \\
   -  \\
   -  \\
\end{matrix}\,\,\,\,\,\,\,\,\begin{matrix}
   0  \\
   0  \\
   |  \\
   |  \\
\end{matrix}\,\,\,\,\,\,\,\begin{matrix}
   {}  \\
   +  \\
   -  \\
   -  \\
\end{matrix}\,\,\,\,\,\begin{matrix}
   5/8  \\
   |  \\
   0  \\
   |  \\
\end{matrix}\,\,\,\,\begin{matrix}
   {}  \\
   +  \\
   +  \\
   -  \\
\end{matrix}\,\,\,\,\,\,\,\begin{matrix}
   5/2  \\
   |  \\
   |  \\
   0  \\
\end{matrix}\,\,\,\,\begin{matrix}
   {}  \\
   +  \\
   +  \\
   +  \\
\end{matrix}\,\,\,\,\,\begin{matrix}
   2  \\
   ||  \\
   ||  \\
   ||  \\
\end{matrix}\,\,\,\,\begin{matrix}
   {}  \\
   +  \\
   +  \\
   +  \\
\end{matrix}\,\,\,\,\,\begin{matrix}
   +\infty   \\
   {}  \\
   {}  \\
   {}  \\
\end{matrix}\,\,\, \\
 & P\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,\,\,||\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,+\,\,\,\,\,||\,\,\,\,\,+\,\,\, \\
\end{align}\]

So the solution is $x\in \left[ 0,\frac{8}{5} \right]\cup \left[ \frac{5}{2},+\infty  \right)$