20. 已知$a+b=3,ab=1$,计算$(a-2)(b-2)$的结果是.

20.-1

三、解答题(共60分)
21. (21分)计算:
(1)$(-3a^{3})^{2}· a^{3}+(-4a)^{2}· a^{7}$;
(2)$(a^{2}· a^{4})^{3}÷ (a^{3})^{2}÷ a$;
(3)$x(x+1)-(x+1)(x-2)$;
(4)$(\dfrac{1}{3})^{-1}+(2023× 2024)^{0}-(-\dfrac{1}{2})^{-2}$;
(5)$[ab(1-a)-2a(b-\dfrac{1}{2})]· (2a^{3}b^{2})$.
(6)$(-2xy)^{2}+(x^{2}y)^{3}÷ (-x^{4}y)$.
(7)$(x+4y)^{2}(x-4y)^{2}$.

21.解:(1)原式$=9a^{6}· a^{3}+16a^{2}· a^{7}=9a^{9}+16a^{9}=25a^{9}.$
(2)原式$=(a^{6})^{3}÷a^{6}÷a=a^{18}÷a^{6}÷a=a^{18-6-1}=a^{11}.$
(3)原式$=x^{2}+x-(x^{2}-2x+x-2)=x^{2}+x-x^{2}+x+2=2x+2.$
(4)原式$=3+1-\frac{1}{(-\frac{1}{2})^{2}}=3+1-4=0.$
(5)原式$=(ab-a^{2}b-2ab+a)· (2a^{3}b^{2})=(-ab-a^{2}b+a)· (2a^{3}b^{2})=-2a^{4}b^{3}-2a^{5}b^{3}+2a^{4}b^{2}.$
(6)原式$=4x^{2}y^{2}+x^{6}y^{3}÷(-x^{4}y)=4x^{2}y^{2}-x^{2}y^{2}=3x^{2}y^{2}.$
(7)原式$=(x^{2}-16y^{2})^{2}=x^{4}-32x^{2}y^{2}+256y^{4}.$