Question

(3x+1)^2-(x+1)^2=0

Answer 1

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

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

=>\(2x\left(4x+2\right)=0\)

=>\(2x\cdot2\left(2x+1\right)=0\)

=>x(2x+1)=0

=>\(\left[{}\begin{matrix}x=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)

Answer 2

`(3x + 1)^2  - (x + 1)^2 = 0`

`=> (3x + 1 - x - 1)(3x + 1 + x + 1) = 0`

`=> 2x(4x + 2) = 0`

TH1:

`2x = 0`

`=> x = 0 : 2 `

`=> x = 0`

TH2:

`4x + 2 = 0`

`=> 4x = -2`

`=>  x = -2/4 = -1/2`

Vây `x = 0 ; x = -1/2`