\(\left|3x^2-3x+1\right|=1-2x\)
ĐKXĐ:1-2x>=0
==>-2x>=-1
==>x<=\(\frac{1}{2}\)
TH1:\(3x^2-3x+1=1-2x\)
\(\Leftrightarrow3x^2-3x+1-1+2x=0\)
\(\Leftrightarrow3x^2-x=0\)
\(\Leftrightarrow x\left(3x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\3x-1=0\end{matrix}\right.=\left\{{}\begin{matrix}x=0\\3x=1\end{matrix}\right.=\left\{{}\begin{matrix}x=0\left(TĐK\right)\\x=\frac{1}{3}\left(TĐK\right)\end{matrix}\right.\)
Vậy ...
TH2 \(3x^2-3x+1=2x-1\)
\(\Leftrightarrow3x^2-3x+1-2x+1=0\)
\(\Leftrightarrow3x^2-5x+2=0\)
\(\Leftrightarrow3x^2-3x-2x+2=0\)
\(\Leftrightarrow(3x^2-3x)-(2x-2)=0\)
\(\Leftrightarrow3x(x-1)-2(x-1)=0\)
\(\Leftrightarrow(3x-2)(x-1)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x-2=0\\x-1=0\end{matrix}\right.=\left\{{}\begin{matrix}3x=2\\x=1\end{matrix}\right.=\left\{{}\begin{matrix}x=\frac{2}{3}\left(KTĐK\right)\\x=1\left(KTĐK\right)\end{matrix}\right.\)
Vậy ...
\(_{\left|3x^2-3x+1\right|=1-2x}\)
3x2-3x+1=1-2x
3x2-x=0
x(3x-1)=0
\(\left\{{}\begin{matrix}x=0\\3x-1=0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x=0\\3x=1\end{matrix}\right.\left\{{}\begin{matrix}x=0\\x=\frac{1}{3}\end{matrix}\right.\)
