\(\left(2x+1\right)^2-\left(3+x\right)^2=0\)
\(\left(2x+1-3-x\right)\left(2x+1+3+x\right)=0\)
\(\left[{}\begin{matrix}x-2=0\\3x+4=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{4}{3}\end{matrix}\right.\)
`(2x+1)^2=(3+x)^2`
`<=>(2x+1)^2-(3+x)^2=0`
`<=>(2x+1-3-x)(2x+1+3+x)=0`
`<=>(x-2)(3x+4)=0`
`<=>` $\left[\begin{matrix} x=2\\ x=\dfrac{-4}{3}\end{matrix}\right.$
Vậy `S={2;[-4]/3}`
\(\left(2x+1\right)^2=\left(3+x\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=3+x\\2x+1=-3-x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\3x=-4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{4}{3}\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{4}{3};2\right\}\)
`=> (2x+1+3+x)(2x+1-3-x) = 0`
`=> (3x + 4)(x - 2) = 0`
`=> 3x + 4 = 0` hoặc `x - 2 =0`
`=> x = -4/3, 2.`