a) ta có : \(\sqrt{4x^2-4x+1}=2\Leftrightarrow\sqrt{\left(2x-1\right)^2}=2\Leftrightarrow\left|2x-1\right|=2\)
th1: \(2x-1\ge0\Leftrightarrow2x\ge\dfrac{1}{2}\) \(\Rightarrow pt\Leftrightarrow2x-1=2\Leftrightarrow x=\dfrac{3}{2}\left(tmđk\right)\)
th2: \(2x-1< 0\Leftrightarrow x< \dfrac{1}{2}\) \(\Rightarrow pt\Leftrightarrow1-2x=2\Leftrightarrow x=\dfrac{-1}{2}\left(tmđk\right)\)
vậy \(x=\dfrac{3}{2};x=\dfrac{-1}{2}\)
b) điều kiện xác định : \(x\left(x+1\right)\ge0\Leftrightarrow\left[{}\begin{matrix}x\ge0\\x\le-1\end{matrix}\right.\)
ta có : \(\sqrt{x\left(x+1\right)}-\sqrt{x^2}=1\Leftrightarrow\sqrt{x\left(x+1\right)}=\sqrt{x^2}+1\)
th1: \(x\ge0\) \(\Rightarrow pt\Leftrightarrow\sqrt{x\left(x+1\right)}=x+1\Leftrightarrow x\left(x+1\right)=x^2+2x+1\)
\(\Leftrightarrow x^2+x=x^2+2x+1\Leftrightarrow x=-1\left(loại\right)\)
th2: \(x\le-1\)
\(\Rightarrow pt\Leftrightarrow\sqrt{x\left(x+1\right)}=1-x\Leftrightarrow x\left(x+1\right)=x^2-2x+1\)\(\Leftrightarrow x^2+x=x^2-2x+1\Leftrightarrow x=\dfrac{1}{3}\left(loại\right)\)
vậy phương trình vô nghiệm .
a. \(\sqrt{4x^2-4x+1}\) = 2
↔ \(\sqrt{\left(2x\right)^2-4x+1}\) = 2
↔ \(\sqrt{\left(2x-1\right)^2}\) = 2
↔\(\left|2x-1\right|\) = 2
↔ \(\left[{}\begin{matrix}2x-1=2\\2x-1=-2\end{matrix}\right.\)
↔ \(\left[{}\begin{matrix}2x=3\\2x=-1\end{matrix}\right.\)
↔ \(\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{-1}{2}\end{matrix}\right.\)
↔ \(\left[{}\begin{matrix}x=1,5\\x=-0,5\end{matrix}\right.\)
Vậy S = \(\left\{1,5;-0,5\right\}\)
