Question

化简求值:(2a+b)^2-(a+1-b)(a+1+b)+(a+1)^2,其中a=2分之1;,b=-2

Answer 1

原式=(4a^2+4ab+b^2)-[(a+1)^2-b^2]+(a+1)^2

=4a^2+4ab+b^2-(a+1)^2+b^2+(a+1)^2

=4a^2+4ab+2b^2

=1-4+8

=5