a: \(2x\left(x+1\right)-2023=\left(x-1\right)^2+x^2\)
=>\(2x^2+2x-2023=x^2-2x+1+x^2\)
=>2x-2023=-2x+1
=>4x=2024
=>\(x=\dfrac{2024}{4}=506\)
b: \(\left(2x-1\right)^2=\dfrac{1}{4}x^2\)
=>\(\left(2x-1\right)^2-\left(\dfrac{1}{2}x\right)^2=0\)
=>\(\left(2x-1-\dfrac{1}{2}x\right)\left(2x-1+\dfrac{1}{2}x\right)=0\)
=>\(\left(\dfrac{3}{2}x-1\right)\left(\dfrac{5}{2}x-1\right)=0\)
=>\(\left[{}\begin{matrix}\dfrac{3}{2}x-1=0\\\dfrac{5}{2}x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{2}x=1\\\dfrac{5}{2}x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1:\dfrac{3}{2}=\dfrac{2}{3}\\x=1:\dfrac{5}{2}=\dfrac{2}{5}\end{matrix}\right.\)
c: \(\left(3x-1\right)x-\dfrac{1}{2}\left(-3x+1\right)=0\)
=>\(x\left(3x-1\right)+\dfrac{1}{2}\left(3x-1\right)=0\)
=>\(\left(3x-1\right)\left(x+\dfrac{1}{2}\right)=0\)
=>\(\left[{}\begin{matrix}3x-1=0\\x+\dfrac{1}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
d: \(10x^2-9x+2=0\)
=>\(10x^2-4x-5x+2=0\)
=>\(2x\left(5x-2\right)-\left(5x-2\right)=0\)
=>(5x-2)(2x-1)=0
=>\(\left[{}\begin{matrix}5x-2=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=\dfrac{1}{2}\end{matrix}\right.\)
