Question

若规定x&y=(x+1)(y+4)-(x+2)(y+3),请化简(m+1)&(n-1),并求m=2008,n=2007时

Answer 1

(m+1)&(n-1),即x=m+1,y=n-1

所以(m+1)&(n-1)

=(m+1+1)(n-1+4)-(m+1+2)(n-1+3)

=(m+2)(n+3)-(m+3)(n+2)

=mn+3m-2n+6-mn-2m-3n-6

=m-n

m=2008,n=2007

(m+1)&(n-1)=2008-2007=1