规定一种新运算:a*b=a+b,aXb=a-b,其中a,b为有理数,化简a^2b*3ab+5a^2bX4ab,并求出当a=5,b=3时的值是多少

问题描述:

规定一种新运算:a*b=a+b,aXb=a-b,其中a,b为有理数,化简a^2b*3ab+5a^2bX4ab,并求出当a=5,b=3时的值是多少

a^2b * 3ab+5a^2b X 4ab
=[(a^2b)+(3ab) ]+ [(5a^2b) - (4ab)]
= ab(a+3)+ab(5a-4)
=ab(a+3+5a-4)
=ab(6a-1)

化简a^2b+3ab+5a^2b-4ab=6a^2b-ab=ab(6a-1)
值为5*3(6*5-1)=435