\(\left(x-3\right)^2=\left(2x+1\right)^2\)
\(\Leftrightarrow\left(x-3\right)^2-\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left(x-3-2x-1\right)\left(x-3-2x+1\right)=0\)
\(\Leftrightarrow\left(-x-4\right)\left(-x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-2\end{matrix}\right.\)
Vậy...
\(\left(x-3\right)^2=\left(2x+1\right)^2\)
\(\Leftrightarrow\left(x-3\right)^2-\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left(x-3-2x-1\right)\left(x-3+2x+1\right)=0\)
\(\Leftrightarrow\left(-x-4\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy....
\(\left(x-3\right)^2=\left(2x+1\right)^2\)
\(\Leftrightarrow x-3=2x+1\)
\(\Leftrightarrow x-2x=3+1\)
\(\Leftrightarrow-x=4\)
\(\Leftrightarrow x=-4\)
Vậy tập nghiệm của pt là : \(S=\left\{-4\right\}\)
