Ta có \(\left|2x-1\right|=\dfrac{1}{2}\)
\(\left|2x-1\right|=2x-1\) khi \(2x-1\ge0\Leftrightarrow2x\ge1\Leftrightarrow x\ge\dfrac{1}{2}\)
\(\left|2x-1\right|=-\left(2x-1\right)\) khi \(2x-1< 0\Leftrightarrow2x< 1\Leftrightarrow x< \dfrac{1}{2}\)
Ta giải hai phương trình sau:
pt1: \(2x-1=\dfrac{1}{2}\left(ĐK:x\ge\dfrac{1}{2}\right)\)
\(\Leftrightarrow2x=\dfrac{1}{2}-1\)
\(\Leftrightarrow2x=-\dfrac{1}{2}\)
\(\Leftrightarrow x=\dfrac{-\dfrac{1}{2}}{2}=-\dfrac{1}{2}\left(ktm\right)\)
pt2: \(-\left(2x-1\right)=\dfrac{1}{2}\left(ĐK:x< \dfrac{1}{2}\right)\)
\(\Leftrightarrow-2x+1=\dfrac{1}{2}\)
\(\Leftrightarrow-2x=\dfrac{1}{2}-1\)
\(\Leftrightarrow-2x=-\dfrac{1}{2}\)
\(\Leftrightarrow x=\dfrac{-\dfrac{1}{2}}{-2}=\dfrac{1}{4}\left(tm\right)\)
Vậy giá trị của đa thức \(E\left(x\right)\) tại \(x=\dfrac{1}{4}\)
\(E\left(x\right)=-4x^4+x+1=-4.\left(\dfrac{1}{4}\right)^4+\dfrac{1}{4}+1=\dfrac{79}{64}\)