Question

若代数式(2x^2+ax-y+6)-(2bx^2-3x+5y-1)的值与x无关,求代数式3(a^2-2ab-b^2)-(

Answer 1

(2x^2+ax-y+6)-(2bx^2-3x+5y-1)

=(2-2b)x^2+(a+3)x-6y+7

所以

2-2b=0

a+3=0

即

b=1

a=-3

3(a^2-2ab-b^2)-(4a^2+ab+b^2)

=-a^2-7ab-4b^2

=-(-3)^2-7*(-3)*1-4*1^2

=8