Question

Cho biểu thức: A = \(\dfrac{\sqrt{4x^2-4x+1}}{4x-2}\). Chứng tỏ | A | = 0,5 với x \(\ne\) 0,5

Answer 1

\(A=\dfrac{\sqrt{4x^2-4x+1}}{4x-2}=\dfrac{\sqrt{\left(2x-1\right)^2}}{2\left(2x-1\right)}=\dfrac{\left|2x-1\right|}{2\left(2x-1\right)}\)

\(\Rightarrow\left|A\right|=\left|\dfrac{\left|2x-1\right|}{2\left(2x-1\right)}\right|=\dfrac{\left|2x-1\right|}{2\left|2x-1\right|}=\dfrac{1}{2}\)

Answer 2

Ta có: \(A=\dfrac{\sqrt{4x^2-4x+1}}{4x-2}\)

\(=\dfrac{\left|2x-1\right|}{2\left(2x-1\right)}\)

\(=\left[{}\begin{matrix}-\dfrac{\left(2x-1\right)}{2\left(2x-1\right)}=-\dfrac{1}{2}\left(x< \dfrac{1}{2}\right)\\\dfrac{2x-1}{2\left(2x-1\right)}=\dfrac{1}{2}\left(x\ge\dfrac{1}{2}\right)\end{matrix}\right.\)

\(\Leftrightarrow\left|A\right|=0.5\)