Ta có: \(A = \dfrac{{\sqrt {4{x^2} - 4x + 1}}}{{4x - 2}}\)\( =\dfrac{{\sqrt {{{\left( {2x - 1} \right)}^2}} }}{{4x - 2}} = \dfrac{{\left| {2x - 1} \right|}}{{2\left( {2x - 1} \right)}}\)
+) Nếu : \(\eqalign{& 2x - 1 > 0 \Leftrightarrow 2x > 1 \cr & \Leftrightarrow x > 0,5 \cr} \)
Suy ra: \(\left| {2x - 1} \right| = 2x - 1\)
Ta có:\(\dfrac{{\left| {2x - 1} \right|}}{{2(2x - 1)}} = \dfrac{{2x - 1}}{{2\left( {2x - 1} \right)}} = \dfrac{1}{2} = 0,5\)
+) Nếu:\(\eqalign{
& 2x - 1 < 0 \Leftrightarrow 2x < 1 \cr
& \Leftrightarrow x < 0,5 \cr} \)
Suy ra: \(\left| {2x - 1} \right| = - (2x - 1)\)
Ta có:
\(\eqalign{
& A = {{\left| {2x - 1} \right|} \over {2\left( {2x - 1} \right)}} = {{ - \left( {2x - 1} \right)} \over {2\left( {2x - 1} \right)}} \cr &=- {1 \over 2} = - 0,5 \cr
& \Rightarrow \left| A \right| = \left| { - 0,5} \right| = 0,5 \cr} \)
Vậy \(|A|=0,5\) với \(x\ne 0,5.\)
