Question

Answer 1

b: Ta có: \(\sqrt{4x^2-4x+1}=\sqrt{x^2-6x+9}\)

\(\Leftrightarrow\left|2x-1\right|=\left|x-3\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=x-3\\2x-1=3-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x-x=-3+1\\2x+x=3+1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{4}{3}\end{matrix}\right.\)

Answer 2

c: Ta có: \(\sqrt{2x^2-2x+1}=2x-1\)

\(\Leftrightarrow4x^2-4x+1-2x^2+2x-1=0\)

\(\Leftrightarrow2x^2-2x=0\)

\(\Leftrightarrow2x\left(x-1\right)=0\)

hay x=1