《2025年轻松作业本八年级数学上册人教版》

13. 计算:
(1)$(4x+3y)(4x-3y)$;
(2)$(-2x-6y)(2x-6y)$;
(3)$(\frac {1}{3}a^{2}-b^{2})(\frac {1}{3}a^{2}+b^{2})$;
(4)$(x+1)(x-1)(x^{2}+1)$.

答案：
(1) $ 16 x^{2}-9 y^{2} $
(2) $ 36 y^{2}-4 x^{2} $
(3) $ \frac{1}{9} a^{4}-b^{4} $
(4) $ x^{4}-1 $

14. (2024·广州天河区二模)已知$T= (2a+3b)(2a-3b)-a(3a-b)+9b^{2}$.
(1) 化简T;
(2) 若a、b互为相反数,求T的值.

答案： 解:
(1) $ T=(2 a+3 b)(2 a-3 b)-a(3 a-b)+9 b^{2}=4 a^{2}-9 b^{2}-3 a^{2}+a b+9 b^{2}=a^{2}+a b $;
(2) $ \because a 、 b $ 互为相反数, $ \therefore a+b=0, \therefore T=a^{2}+a b=a(a+b)=0 $.

15. 学完平方差公式后,王老师展示了以下例题:
例:计算$(1+\frac {1}{2})(1+\frac {1}{2^{2}})(1+\frac {1}{2^{4}})(1+\frac {1}{2^{8}})+\frac {1}{2^{15}}$.
观察算式发现:如果将$(1+\frac {1}{2})乘(1-\frac {1}{2})$,这时可以连续运用平方差公式进行计算,为使等式恒成立,需将式子整体再乘2.
解:原式$=2×(1-\frac {1}{2})(1+\frac {1}{2})(1+\frac {1}{2^{2}})(1+\frac {1}{2^{4}})(1+\frac {1}{2^{8}})+\frac {1}{2^{15}}$
$=2×(1-\frac {1}{2^{2}})(1+\frac {1}{2^{2}})(1+\frac {1}{2^{4}})(1+\frac {1}{2^{8}})+\frac {1}{2^{15}}$
$=2×(1-\frac {1}{2^{4}})(1+\frac {1}{2^{4}})(1+\frac {1}{2^{8}})+\frac {1}{2^{15}}$
$=2×(1-\frac {1}{2^{8}})(1+\frac {1}{2^{8}})+\frac {1}{2^{15}}$
$=2×(1-\frac {1}{2^{16}})+\frac {1}{2^{15}}$
$=2-\frac {1}{2^{15}}+\frac {1}{2^{15}}$
$=2$.
(1) 请仿照上述方法计算:$2(3+1)(3^{2}+1)(3^{4}+1)(3^{8}+1)+1$;
(2) 请认真观察,计算:$(1-\frac {1}{2^{2}})(1-\frac {1}{3^{2}})(1-\frac {1}{4^{2}})(1-\frac {1}{5^{2}})$.
(1) $(3-1)(3+1)(3^{2}+1)(3^{4}+1)(3^{8}+1)+1=(3^{2}-1)(3^{2}+1)(3^{4}+1)(3^{8}+1)+1=(3^{4}-1)(3^{4}+1)(3^{8}+1)+1=(3^{8}-1)(3^{8}+1)+1=3^{16}-1+1=3^{16}$; (2) $(1+\frac{1}{2})(1-\frac{1}{2})(1+\frac{1}{3})(1-\frac{1}{3})(1+\frac{1}{4})(1-\frac{1}{4})(1+\frac{1}{5})(1-\frac{1}{5})=\frac{3}{2}×\frac{1}{2}×\frac{4}{3}×\frac{2}{3}×\frac{5}{4}×\frac{3}{4}×\frac{6}{5}×\frac{4}{5}=\frac{1}{2}×\frac{6}{5}=\frac{3}{5}$.

答案： 解:
(1) $ 2(3+1)\left(3^{2}+1\right)\left(3^{4}+1\right)\left(3^{8}+1\right)+1=(3-1)(3+1)\left(3^{2}+1\right)\left(3^{4}+1\right)\left(3^{8}+1\right)+1=\left(3^{2}-1\right)\left(3^{2}+1\right)\left(3^{4}+1\right)\left(3^{8}+1\right)+1=\left(3^{4}-1\right)\left(3^{4}+1\right)\left(3^{8}+1\right)+1=\left(3^{8}-1\right)\left(3^{8}+1\right)+1=3^{16}-1+1=3^{16} $;
(2) $ \left(1-\frac{1}{2^{2}}\right)\left(1-\frac{1}{3^{2}}\right)\left(1-\frac{1}{4^{2}}\right)\left(1-\frac{1}{5^{2}}\right)=\left(1+\frac{1}{2}\right)\left(1-\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1-\frac{1}{3}\right)\left(1+\frac{1}{4}\right)\left(1-\frac{1}{4}\right)\left(1+\frac{1}{5}\right)\left(1-\frac{1}{5}\right)=\frac{3}{2} \times \frac{1}{2} \times \frac{4}{3} \times \frac{2}{3} \times \frac{5}{4} \times \frac{3}{4} \times \frac{6}{5} \times \frac{4}{5}=\frac{1}{2} \times \frac{6}{5}=\frac{3}{5} $.