\(\left(4x+3\right)^2=4\left(x^2-2x+1\right)\)
\(\Leftrightarrow\left(4x+3\right)^2=4\left(x-1\right)^2\)
\(\Leftrightarrow\left(4x+3\right)^2-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(4x+3+2x-2\right)\left(4x+3-2x+2\right)=0\)
\(\Leftrightarrow\left(6x+1\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{1}{6}\\x=-\frac{5}{2}\end{matrix}\right.\)
Vậy ...
\(\left(4x+3\right)^2=4x^2-8x+4\)
\(\Leftrightarrow\left(4x+3\right)^2=\left(2x-2\right)^2\)
\(\Leftrightarrow\left(4x+3\right)^2-\left(2x-2\right)^2=0\)
\(\Leftrightarrow\left(2x-5\right)\left(6x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-5=0\\6x+1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=\frac{-1}{6}\end{matrix}\right.\)
