Question

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

giải phương trình nhé

Answer 1

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

`<=> ( 2x + 3 - x - 1 ) ( 2x + 3 + x + 1 ) = 0`

`<=> ( x + 2 ) ( 3x + 4 ) = 0`

`<=>` $\left[\begin{matrix} x + 2 = 0\\ 3x + 4 = 0\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x=-2\\ x=\dfrac{-4}{3}\end{matrix}\right.$

Vậy `S = { -2 ; [-4] / 3 }`

Answer 2

=>(2x+3-x-1).(2x+3+x+1)=0

=>(x+2).(3x+4)=0

=>x+2=0 =>x=-2

hoặc 3x+4=0=>x=-4/3