\(\left(x+2\right)^2-\left(x+2\right)\left(x-2\right)\\ =\left(x+2\right)\left(x+2-x+2\right)\\ =4\left(x+2\right)=4x+8\)
Ta có:(x+2)2-(x+2)(x-2)=(x+2)(x+2-x+2)=4(x+2)
`(x+2)^2-(x+2)(x-2)`
`=x^2+4x+4-(x^2-4)`
`=x^2+4x+4-x^2+4`
`=4x+8`
\(\left(x+2\right)^2-\left(x+2\right)\left(x-2\right)\)
\(=\left(x+2\right)\left(x+2-x+2\right)\)
=4x+8