\(\left(2x+5\right)^2=\left(x+2\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+5=x+2\\2x+5=-x-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-x=2-5\\2x+x=-2-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{7}{3}\end{matrix}\right.\)
easy thôi
\(\left(2x+5\right)^2=\left(x+2\right)^2\)
\(\Leftrightarrow\left(2x+5\right)^2-\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(2x+5-x-2\right)\left(2x+5+x+2\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(3x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\3x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{7}{3}\end{matrix}\right.\)
Vậy \(S=\left\{-3;-\dfrac{7}{3}\right\}\)
\(\left(2x+5\right)^2=\left(x+2\right)^2\)
<=> \(\left(2x+5\right)^2-\left(x+2\right)^2=0\)
<=> \(\left[{}\begin{matrix}2x+5-x-2=0\\2x+5+x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-3=0\\3x+7=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=3\\x=\dfrac{-7}{3}\end{matrix}\right.\)
\(\left(2x+5\right)^2=\left(x+2\right)^2\)
\(\Leftrightarrow\left(2x+5\right)^2-\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(2x+5-x-2\right)\left(2x+5+x+2\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(3x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{7}{3}\end{matrix}\right.\)