\(x+\sqrt{x^2-4x+4}=\dfrac{1}{2}\\ \Rightarrow\sqrt{x^2-4x+4}=\dfrac{1}{2}-x\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2}-x\ge0\\x^2-4x+4=\left(\dfrac{1}{2}-x\right)^2\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x\le\dfrac{1}{2}\\x^2-4x+4=x^2-x+\dfrac{1}{4}\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x\le\dfrac{1}{2}\\3x=\dfrac{15}{4}\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x\le\dfrac{1}{2}\\x=\dfrac{5}{4}\end{matrix}\right.\left(L\right)\)
Vậy PT vô nghiệm
=>|x-2|=1/2-x
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{1}{2}\\\left(x-2-\dfrac{1}{2}+x\right)\left(x-2+\dfrac{1}{2}-x\right)=0\end{matrix}\right.\)
\(\Leftrightarrow x\in\varnothing\)
