《2025年一遍过八年级数学上册人教版》

1 若$-2x^{2}(3x^{2}-ax-6)-3x^{3}+x^{2}$中不含x的三次项,则$a= $____.

答案： $\frac{3}{2}$ $-2x^{2}(3x^{2}-ax-6)-3x^{3}+x^{2}=-6x^{4}+2ax^{3}+12x^{2}-3x^{3}+x^{2}=-6x^{4}+(2a-3)x^{3}+13x^{2}$.因为$-2x^{2}(3x^{2}-ax-6)-3x^{3}+x^{2}$中不含x的三次项,所以$2a-3=0$,解得$a=\frac{3}{2}.$

2 计算下列各题:
(1)$(a+1)(2-b)-a(1-b)-2;$
(2)$[x(x^{2}y^{2}-xy)-y(x^{2}-x^{3}y)]÷(x^{2}y).$

答案： 解:
(1)$(a+1)(2-b)-a(1-b)-2$ $=2a-ab+2-b-a+ab-2$ $=a-b.$
(2)$[x(x^{2}y^{2}-xy)-y(x^{2}-x^{3}y)]÷(x^{2}y)$ $=(x^{3}y^{2}-x^{2}y-x^{2}y+x^{3}y^{2})÷(x^{2}y)$ $=(2x^{3}y^{2}-2x^{2}y)÷(x^{2}y)$ $=2xy-2.$

3 计算:
(1)$(x+y-1)(x+y+1)-(x-2y)(x+2y);$
(2)$(x+y)^{2}(x-y)^{2};$
(3)$(a-b)(a+b)(a^{2}+b^{2})(a^{4}+b^{4})(a^{8}+b^{8})(a^{16}+b^{16});$
变式$(1+4^{2})(1+4^{4})(1+4^{8})(1+4^{16}).$

答案： 解:
(1)$(x+y-1)(x+y+1)-(x-2y)(x+2y)$ $=(x+y)^{2}-1-(x^{2}-4y^{2})$ $=x^{2}+2xy+y^{2}-1-x^{2}+4y^{2}$ $=5y^{2}+2xy-1.$
(2)$(x+y)^{2}(x-y)^{2}$ $=[(x+y)(x-y)]^{2}$ $=(x^{2}-y^{2})^{2}$ $=x^{4}-2x^{2}y^{2}+y^{4}.$
(3)$(a-b)(a+b)(a^{2}+b^{2})(a^{4}+b^{4})(a^{8}+b^{8})(a^{16}+b^{16})$ $=(a^{2}-b^{2})(a^{2}+b^{2})(a^{4}+b^{4})(a^{8}+b^{8})(a^{16}+b^{16})$ $=(a^{4}-b^{4})(a^{4}+b^{4})(a^{8}+b^{8})(a^{16}+b^{16})$ $=(a^{8}-b^{8})(a^{8}+b^{8})(a^{16}+b^{16})$ $=(a^{16}-b^{16})(a^{16}+b^{16})$ $=a^{32}-b^{32}.$ 变式 $(1+4^{2})(1+4^{4})(1+4^{8})(1+4^{16})$ $=\frac{1}{15}(4^{2}-1)(1+4^{2})(1+4^{4})(1+4^{8})(1+4^{16})$ $=\frac{1}{15}(4^{4}-1)(1+4^{4})(1+4^{8})(1+4^{16})$ $=\frac{1}{15}(4^{8}-1)(1+4^{8})(1+4^{16})$ $=\frac{1}{15}(4^{16}-1)(1+4^{16})$ $=\frac{1}{15}(4^{32}-1).$

4 先化简,再求值:$(x+2)(3x-2)-2x(x+2)$,其中$x= 3.$

答案： 解:$(x+2)(3x-2)-2x(x+2)$ $=3x^{2}-2x+6x-4-2x^{2}-4x$ $=x^{2}-4.$ 当$x=3$时,原式$=3^{2}-4=5.$

5 先化简,再求值:$[(3a+b)^{2}+(b+3a)(b-3a)-6b^{2}]÷(2b)$,其中$a= -\frac {1}{3},b= -2.$

答案： 解:$[(3a+b)^{2}+(b+3a)(b-3a)-6b^{2}]÷(2b)$ $=(9a^{2}+6ab+b^{2}+b^{2}-9a^{2}-6b^{2})÷(2b)$ $=(-4b^{2}+6ab)÷(2b)$ $=-2b+3a.$ 当$a=-\frac{1}{3},b=-2$时,原式$=-2×(-2)+3×(-\frac{1}{3})=4-1=3.$