(x+1)2-(2x-1)2+3(x-2)(x+2)
=x2+2x+1-4x2+4x-1+3(x2-4)
=x2+2x+1-4x2+4x-1+3x2-12
= 6x-12
= 6(x-2)
=6(-1-2)
=6.-3
=-18
Ta có: \(\left(x+1\right)^2-\left(2x-1\right)^2+3\left(x-2\right)\left(x+2\right)\)
\(=x^2+2x+1-4x^2+4x-1+3x^2-12\)
\(=6x-12\)(1)
Thay x=1 vào (1), ta được:
\(6\cdot1-12=6-12=-6\)