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any short way to solve this . \(\large \color{black}{\begin{align} \left|\dfrac{x^2-5x+4}{x^2-4}\right|\leq 1\hspace{.33em}\\~\\ \end{align}}\)
\(\large \left|\dfrac{x^2-5x+4}{x^2-4}\right| = \left|\dfrac{(x^2-4)-5x+8}{x^2-4}\right| = \left|1+\dfrac{-5x+8}{x^2-4}\right|\) For this to be \(\le 1\), clearly we must have \[-2\lt\dfrac{-5x+8}{x^2-4}\lt 0\]
u mean \(\large \color{black}{\begin{align} -2\leq\dfrac{-5x+8}{x^2-4}\leq 0\hspace{.33em}\\~\\ \end{align}}\) ?

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i am also confused on how to further solve this . \(\large \color{black}{\begin{align} -2\leq\dfrac{-5x+8}{x^2-4}\leq 0\hspace{.33em}\\~\\ \end{align}}\)
Ahh yes, consider two cases : Case1 : \(x^2-4\gt 0\) multiply \(x^2-4\) through out and get \[-2(x^2-4 )\le -5x+8 \le 0 \\~\\\implies -2x^2 \le -5x \le -8 \\~\\\implies 2x^2\ge 5x\ge 8 \\~\\\implies x\ge \frac{5}{2} \land x\ge \frac{8}{5}\\~\\\implies x\ge \frac{5}{2} \]
similarly you can work the other case i don't like this method yeah too boring
\[\large \color{black}{\begin{align} \left|\dfrac{x^2-5x+4}{x^2-4}\right|\leq 1\hspace{.33em}\\~\\ \end{align}}\] Since we have the absolute value, we need to consider two cases. One when we have an all positive problem like this \[\frac{x^2-5x+4}{x^2-4} \leq 1 \] or \[-(\frac{x^2-5x+4}{x^2-4}) \leq 1 \] I'm going to try the all positive case \[\frac{x^2-5x+4}{x^2-4} \leq 1 \] so I would distribute the \[x^2-4 \] to the other side \[x^2-5x+4 \leq 1(x^2-4)\] \[x^2-5x+4 \leq x^2-4 \] \[x^2-5x+4 -x^2+4 \leq 0\] (and then put everything on the left... combine like terms) \[x^2-x^2-5x+4 +4 \leq 0\] \[-5x+8 \leq 0\] (simplify and divide. since we have a negative number, the sign switches from \[\leq \] to \[\geq \] \[-5x\leq -8\] \[x\geq \frac{8}{5}\]
there has got to be a way to make Latex nicer. this is ugly :P
the above needs to consider this two cases also \(\dfrac{x^2-5x+4}{x^2-4}\geq 0,\ \ \dfrac{x^2-5x+4}{x^2-4}< 0\)
I said I just picked one case :P I'm aware that there are two.
is there a way to transform the question in to this. \(\large \color{black}{\begin{align} \left|\dfrac{x^2-5x+4}{x^2-4}+\color{black}{constant}\right|\leq \color{red}{0}\hspace{.33em}\\~\\ \end{align}}\)
maybe subtract that 1 on both sides? But I have never done it, so I don't know if it's a legal math move.
i m not saying substract \(1\) lol
\[\frac{-x^2+5x-4}{x^2-4} \leq 1\] \[-x^2+5x-4 \leq 1(x^2-4)\] \[-x^2+5x-4 \leq x^2-4\] \[-x^2-x^2+5x-4+4 \leq 0\] \[-2x^2+5x\leq 0\] \[x(-2x+5)\leq 0\] x would be 0... blah don't want that case \[-2x+5 \leq 0 \] \[-2x \leq -5 \] \[x\geq \frac{5}{2} \] Looks like there is no faster way of doing this. You just have to use the absolute value rules which is consider the cases where we have an all positive situation and one where the negative sign is being distributed.
Why you have chosen 1 to subtract @UsukiDoll ?
We have two cases thanks to the absolute value remember we either have all positive or we have a negative sign attached to the problem and have to distribute it and then do the same process as above. @ganeshie8 had the same answers I have.
\(\color{blue}{\text{Originally Posted by}}\) @waterineyes Why you have chosen 1 to subtract @UsukiDoll ? \(\color{blue}{\text{End of Quote}}\) didn't do it.. just in case that move didn't make sense :P
Which move?

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