given 1<x<3., Find the value of \(A=\frac{\left|x-3\right|}{x-3}-\frac{\left|x-1\right|}{1-x} +\left|x-1\right|+\left|3-x\right|\)
Answer:A=...........
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Lời giải:
Vì \(1< x< 3\Rightarrow \left\{\begin{matrix} |x-3|=|3-x|=3-x\\ |x-1|=x-1\end{matrix}\right.\). Khi đó:
\(A=\frac{|x-3|}{x-3}-\frac{|x-1|}{1-x}+|x-1|+|3-x|\)
\(=\frac{3-x}{x-3}-\frac{x-1}{1-x}+x-1+3-x\)
\(=-1-(-1)+2=2\)
Vậy giá trị của $A$ là $2$