- Biết a – b = 7 tính : A = a2(a + 1) – b2(b – 1) + ab – 3ab(a – b + 1)
Ta có:
$a^2(a+1)-b^2(b-1)+ab-3ab(a-b+1)$
=$a^3+a^2-b^3+b^2+ab-3a^2b+3ab^2-3ab$
=$a^3-3a^2b+3ab^2-b^3+a^2-2ab+b^2$
=$(a-b)^3+(a-b)^2$
Vì a-b=7(gt)
nên $(a-b)^3+(a-b)^2=7^3+7^2=392$
Vậy $a^2(a+1)-b^2(b-1)+ab-3ab(a-b+1)=392 khi a-b=7$