\(\sqrt{2x^2-2x+1}=2x-1\)
\(\Leftrightarrow2x^2-2x+1=4x^2-4x+1\)
\(\Leftrightarrow2x^2-2x=0\)
\(\Leftrightarrow2x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\)
Vậy S=\(\left\{0,1\right\}\)
ĐKXĐ:
\(\left\{{}\begin{matrix}2x^2-2x+1\ge0\\2x-1\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2+x^2\ge0\forall x\\x\ge\dfrac{1}{2}\end{matrix}\right.\)
Đề bài
\(\Rightarrow2x^2-2x+1=\left(2x-1\right)^2=4x^2-4x+1\)
\(\Leftrightarrow2x^2-2x=0\)
\(\Leftrightarrow2x\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=1\left(tm\right)\end{matrix}\right.\)
Vậy pt có tập nghiệm \(S=\left\{1\right\}\)
Ta có: \(\sqrt{2x^2-2x+1}=2x-1\)
\(\Leftrightarrow4x^2-4x+1-2x^2+2x-1=0\)
\(\Leftrightarrow2x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)
