Question

tìm x, biết:

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

Answer 1

`(2x+3)(3-2x)+(2x-1)^{2}=2`

`<=>3^{2}-(2x)^{2}+(2x)^{2}-2.2x.1+1^{2}=2`

`<=>9-4x^{2}+4x^{2}-4x+1=2`

`<=>-4x+10=2`

`<=>-4x=-8`

`<=>x=2`

Answer 2

\(6x+9-4x^2-6x+4x^2-4x+1-2=0\\ -4x+8=0\\ -4x=-8\\ x=2\)

Answer 3

      (2x+3).(3-2x)+(2x-1)^2=2

<=>9-4x^2+4x^2-4x+1=2

<=>-4x+10=2

<=>-4x=-8

<=>x=2

Vậy x=2