零五网 全部参考答案 启东中学作业本 2025年启东中学作业本八年级数学上册人教版 第16页解析答案
1. 把下列各式因式分解:
(1)$x^{2}-9$; (2)$16x^{2}-9y^{2}$;
(3)$(y+1)^{2}-9$; (4)$m^{2}-(2m+3)^{2}$;
(5)$(x^{2}+4y^{2})^{2}-16x^{2}y^{2}$; (6)$4(a-b)^{2}-(a+b)^{2}$;
(7)$(m+1)(m-9)+8m$; (8)$9a^{2}-4(b+c)^{2}$;
(9)$-x^{4}+16$; (10)$2x^{2}-\frac {1}{2}y^{2}$.
答案:(1)解:$x^{2}-9=(x+3)(x-3)$
(2)解:$16x^{2}-9y^{2}=(4x)^{2}-(3y)^{2}=(4x+3y)(4x-3y)$
(3)解:$(y+1)^{2}-9=(y+1)^{2}-3^{2}=(y+1+3)(y+1-3)=(y+4)(y-2)$
(4)解:$m^{2}-(2m+3)^{2}=[m+(2m+3)][m-(2m+3)]=(3m+3)(-m-3)=-3(m+1)(m+3)$
(5)解:$(x^{2}+4y^{2})^{2}-16x^{2}y^{2}=(x^{2}+4y^{2})^{2}-(4xy)^{2}=(x^{2}+4y^{2}+4xy)(x^{2}+4y^{2}-4xy)=(x+2y)^{2}(x-2y)^{2}$
(6)解:$4(a-b)^{2}-(a+b)^{2}=[2(a-b)]^{2}-(a+b)^{2}=[2(a-b)+(a+b)][2(a-b)-(a+b)]=(2a-2b+a+b)(2a-2b-a-b)=(3a-b)(a-3b)$
(7)解:$(m+1)(m-9)+8m=m^{2}-9m+m-9+8m=m^{2}-9=(m+3)(m-3)$
(8)解:$9a^{2}-4(b+c)^{2}=(3a)^{2}-[2(b+c)]^{2}=(3a+2(b+c))(3a-2(b+c))=(3a+2b+2c)(3a-2b-2c)$
(9)解:$-x^{4}+16=16-x^{4}=4^{2}-(x^{2})^{2}=(4+x^{2})(4-x^{2})=(4+x^{2})(2+x)(2-x)$
(10)解:$2x^{2}-\frac{1}{2}y^{2}=\frac{1}{2}(4x^{2}-y^{2})=\frac{1}{2}[(2x)^{2}-y^{2}]=\frac{1}{2}(2x+y)(2x-y)$
解析:
 
2. 在实数范围内分解因式:[提示:$2= (\sqrt {2})^{2}$]
(1)$x^{2}-3$; (2)$5x^{2}-3$.
答案:2.(1)(x+$\sqrt{3}$)(x-$\sqrt{3}$) (2)($\sqrt{5}$x+$\sqrt{3}$)($\sqrt{5}$x-$\sqrt{3}$)
解析:
(1)解:原式$=x^{2}-(\sqrt{3})^{2}=(x+\sqrt{3})(x-\sqrt{3})$
(2)解:原式$=(\sqrt{5}x)^{2}-(\sqrt{3})^{2}=(\sqrt{5}x+\sqrt{3})(\sqrt{5}x-\sqrt{3})$
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