\(\dfrac{\left|x-4\right|}{x-4}\)
+ ) \(\left|x-4\right|\Leftrightarrow\left[{}\begin{matrix}x-4\\4-x\end{matrix}\right.\)
TH1 : \(\dfrac{x-4}{x-4}=1\)
TH2 : \(\dfrac{4-x}{x-4}\)
Ta có:\(\left|x-4\right|\Leftrightarrow\left[{}\begin{matrix}x-4\\4-x\end{matrix}\right.\)
\(\Rightarrow\)TH1:\(\dfrac{\left|x-4\right|}{x-4}=\dfrac{x-4}{x-4}=1\)
TH2;\(\dfrac{\left|x-4\right|}{x-4}=\dfrac{4-x}{x-4}=-1\)
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