规定一种新运算:a*b=a+b,a@b=a-b,其中a、b为有理数,化简a^2b*3ab+5a^2b@4ab,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,a@b=a-b,其中a、b为有理数,化简a^2b*3ab+5a^2b@4ab,并求出当a=5,b=3时的值是多
少
答
a^2b*3ab+5a^2b@4ab=(a^2b+3ab)+(5a^2b-4ab)=6a^2b-ab
当a=5,b=3时,
原式=6*25*3-5*3=435
答
a^2b*3ab+5a^2b@4ab
=a^2b+3ab+5a^2b-4ab,
=6a^2b-ab
=ab(6a-1)
=5*3*(6*5-1)
=435