《2025年新课标教材同步导练八年级数学下册北师大版C版》

[例3]因式分解:
(1)$a^{4}-81b^{4}$ (2)$a^{4}-8a^{2}b^{2}+16b^{4}$ (3)$(x^{2}+y^{2})^{2}-4x^{2}y^{2}$ (4)$(x^{2}+1)^{2}-4x(x^{2}+1)+4x^{2}$

答案： 解
(1)$a^{4}-81b^{4}=(a^{2})^{2}-(9b^{2})^{2}=(a^{2}+9b^{2})(a^{2}-9b^{2})=(a^{2}+9b^{2})(a + 3b)(a - 3b)$
(2)$a^{4}-8a^{2}b^{2}+16b^{4}=(a^{2})^{2}-2\cdot a^{2}\cdot4b^{2}+(4b^{2})^{2}$$=(a^{2}-4b^{2})^{2}=[(a + 2b)(a - 2b)]^{2}=(a + 2b)^{2}(a - 2b)^{2}$
(3)$(x^{2}+y^{2})^{2}-4x^{2}y^{2}=(x^{2}+y^{2})^{2}-(2xy)^{2}$$=(x^{2}+y^{2}+2xy)(x^{2}+y^{2}-2xy)=(x + y)^{2}(x - y)^{2}$
(4)$(x^{2}+1)^{2}-4x(x^{2}+1)+4x^{2}=(x^{2}+1)^{2}-2(x^{2}+1)\cdot2x+(2x)^{2}$$=[(x^{2}+1)-2x]^{2}=(x^{2}-2x + 1)^{2}=[(x - 1)^{2}]^{2}=(x - 1)^{4}$

[例4]计算下列各题.
(1)$1.99^{2}-2.99^{2}$； (2)$202^{2}+202\times196 + 98^{2}$.

答案： 解
(1)$1.99^{2}-2.99^{2}=(1.99 + 2.99)(1.99 - 2.99)=4.98\times(-1)=-4.98$.
(2)$202^{2}+202\times196 + 98^{2}=202^{2}+2\times202\times98 + 98^{2}=(202 + 98)^{2}=300^{2}=90000$.