giải phương trình sau (3x+1)^2-16.(x+1)^2=0

`(3x+1)^2-16(x+1)^2=0`

`<=>` `(3x+1)^2-[4(x+1)]^2=0`

`<=>` `(3x+1)^2-(4x+4)^2=0`

`<=>` `(3x+1+4x+4).[3x+1-(4x+4)]=0`

`<=>` `(7x+5).(3x+1-4x-4)=0`

`<=>` `(7x+5)(-x-3)=0`

`<=>` `[(7x+5=0),(-x-3=0):}`

`<=>` `[(x=-5/7),(x=-3):}`

Vậy `S={-5/7;-3}`