Question

giải phương trình

\(\left(3x+1\right)^2=\left(x+2\right)^2\)

Answer 1

\(\Leftrightarrow\left(3x+1\right)^2-\left(x+2\right)^2=0\Leftrightarrow\left(2x-1\right)\left(4x+3\right)=0\Leftrightarrow x=\dfrac{1}{2};x=-\dfrac{3}{4}\)

Answer 2

\(\left(3x+1\right)^2=\left(x+2\right)^2\\ \Leftrightarrow\left[{}\begin{matrix}3x+1=x+2\\3x+1=-x-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=1\\4x=-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{-3}{4}\end{matrix}\right.\)

Answer 3

x=-\(\dfrac{3}{4}\)

Answer 4

x=-

Answer 5

\(\left(3x+1\right)^2=\left(x+2\right)^2\) ⇔ \(\left(3x+1\right)^2-\left(x+2\right)^2=0\) 

⇔\(\left(3x+1+x+2\right)\left(3x+1-x-2\right)=0\) 

⇔\(\left(4x+3\right)\left(2x-1\right)=0\)

⇔\(4x+3=0\) hay \(2x-1=0\)

⇔\(x=\dfrac{-3}{4}\) hay \(x=\dfrac{1}{2}\)