设a=2011x+2012=A-1,则b=2011x+2013=A,c=2011x+2014=A+1,
则多项式a的平方+b的平方+c的平方-ab-bc-ac的值
=(A-1)(A-1)+AA+(A+1)(A+1)-(A-1)A-A(A+1)-(A-1)(A+1)
=AA-2A+1+AA+AA+2A+1-AA+A-AA-A-AA+1
=3
如果答案正确,
设a=2011x+2012=A-1,则b=2011x+2013=A,c=2011x+2014=A+1,
则多项式a的平方+b的平方+c的平方-ab-bc-ac的值
=(A-1)(A-1)+AA+(A+1)(A+1)-(A-1)A-A(A+1)-(A-1)(A+1)
=AA-2A+1+AA+AA+2A+1-AA+A-AA-A-AA+1
=3
如果答案正确,