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