若规定a*b=(a+1)(b+4)-(a+2)(b+3),请化简(m+1)*(n-1),并求当m=2005,n=2004时,式子的值
问题描述:
若规定a*b=(a+1)(b+4)-(a+2)(b+3),请化简(m+1)*(n-1),并求当m=2005,n=2004时,式子的值
答
(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
=2005-2004=1
答
(m+1)*(n-1)
=(m+1+1)(n-1+4)-(m+1+2)(n-1+3)
=(m+2)(n-3)-(m+3)(n+2)
=mm-3m+2n-6-mn-2m-3n-6
=-5m-n-12
=-5*2005-2004-12
=-12041
答
设m+1=a n-1=b(m+1)*(n-1)=a*b=(a+1)(b+4)-(a+2)(b+3)=(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当m=2005,n=2004时.原式=2005-2004=1
答
(m+1)*(n-1)=(m+1+1)(n-1+4)-(m+1+2)(n-1+3)=(m+2)(n+3)-(m+3)(n+2)=m-n
m=2005,n=2004时,式子的值=1