Question

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

Answer 1

\(4x^2+4x+1-4\left(x^2+4x+4\right)=0\\ 4x^2+4x+1-4x^2-16x-16=0\\ -12x-15=0\\ -12x=15\\ x=15:\left(-12\right)\\ x=-\dfrac{15}{12}\)

Answer 2

`(2x+1)^2 - 4.(x+2)^2 = 0`

`4x^2 +4x + 1 -4x^2 -16x - 16 =0 `

`=> -12x - 15 =0`

`=> -12x = 15`

`=> x = 15 : (-12)`

`=> x= -5/4`

Vậy `x = -5/4`

`#LeMichael`

Answer 3

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

\(\Rightarrow4x^2+4x+1-4\left(x^2+4x+4\right)\)

\(\Rightarrow4x^2+4x+1-4x^2-16x-16=0\)

\(\Rightarrow-12x=15\)

\(\Rightarrow x=-\dfrac{5}{4}\)