Question

giải PT, help me!!! Gấp lắm ạ

\(\sqrt{4x^2-4x+1}=x-1\)

Answer 1

ĐKXĐ: \(x\ge1\)

\(pt\Leftrightarrow\sqrt{\left(2x-1\right)^2}=x-1\Leftrightarrow\left|2x-1\right|=x-1\)

\(\Leftrightarrow2x-1=x-1\left(do.x\ge1\right)\)

\(\Leftrightarrow x=0\left(ktm\right)\)

Vậy \(S=\varnothing\)

Answer 2

ĐK \(x\ge1\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=\dfrac{2}{3}\left(ktm\right)\end{matrix}\right.\)