\(D=\dfrac{4x^2-4x+1-2x}{4x^2-4x+1}=1-\dfrac{2x}{\left(2x-1\right)^2}=1-\dfrac{2x-1+1}{\left(2x-1\right)^2}=1-\dfrac{1}{2x-1}-\dfrac{1}{\left(2x-1\right)^2}\)
\(D=\dfrac{5}{4}-\left(\dfrac{1}{\left(2x-1\right)^2}+2.\dfrac{1}{\left(2x-1\right)}.\dfrac{1}{2}+\dfrac{1}{4}\right)=\dfrac{5}{4}-\left(\dfrac{1}{2x-1}+\dfrac{1}{2}\right)^2\le\dfrac{5}{4}\)
\(\Rightarrow D_{max}=\dfrac{5}{4}\) khi \(\dfrac{1}{2x-1}+\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{-1}{2}\)
\(D=\dfrac{4x^2-6x+1}{4x^2-4x+1}=\dfrac{4x^2-4x+1-2x+1-1}{4x^2-4x+1}=\dfrac{\left(2x-1\right)^2-\left(2x-1\right)-1}{\left(2x-1\right)^2}=1-\dfrac{1}{2x-1}-\dfrac{1}{\left(2x-1\right)^2}\)
Đặt \(\dfrac{1}{2x-1}=a\) , ta có :
\(D=1-a-a^2=-\left(a^2+a+\dfrac{1}{4}\right)+\dfrac{5}{4}=\dfrac{5}{4}-\left(a+\dfrac{1}{2}\right)^2\le\dfrac{5}{4}\)
Dấu " = " xảy ra \(\Leftrightarrow a=-\dfrac{1}{2}\Leftrightarrow\dfrac{1}{2x-1}=-\dfrac{1}{2}\Leftrightarrow2x-1=-2\Leftrightarrow x=-\dfrac{1}{2}\)
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