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八年级数学填空题一般
题目
数学活动课上,老师准备了若干个如图11的三种纸片,AA种纸片是边长为aa的正方形,BB种纸片是边长为bb的正方形,CC种纸片是长为bb、宽为aa的长方形,并用AA种纸片一张,BB种纸片一张,CC种纸片两张拼成如图22的大正方形.
(1)(1)观察图22,请你写出下列三个代数式:(a+b)2\left(a+b\right)^{2},a2+b2a^{2}+b^{2},abab之间的等量关系.
(2)(2)若要拼出一个面积为(a+2b)(a+b)\left(a+2b\right)\left(a+b\right)的矩形,则需要AA号卡片11张,BB号卡片22张,CC号卡片______张.
(3)(3)根据(1)题中的等量关系,解决如下问题:
①已知:a+b=5a+b=5,a2+b2=11a^{2}+b^{2}=11,求abab的值;
②已知(x2021)2+(x2023)2=20\left(x-2021\right)^{2}+\left(x-2023\right)^{2}=20,求x2022x-2022的值.
知识点:多项式乘多项式、完全平方公式的几何背景章节:第16章 整式的乘法 / 16.3 乘法公式 / 16.3.2 完全平方公式

答案与解析

答案

(1)大正方形的面积可以表示为:(a+b)2\left(a+b\right)^{2},或表示为:a2+b2+2aba^{2}+b^{2}+2ab
因此有(a+b)2=a2+b2+2ab\left(a+b\right)^{2}=a^{2}+b^{2}+2ab
(2)(2)根据题意得:(a+2b)(a+b)=a2+3ab+2b2\left(a+2b\right)\left(a+b\right)=a^{2}+3ab+2b^{2}
则需要AA号卡片11张,BB号卡片22张,CC号卡片33张.
故答案为:33
(3)(3)(a+b)2=a2+b2+2ab\because \left(a+b\right)^{2}=a^{2}+b^{2}+2aba+b=5a+b=5a2+b2=11a^{2}+b^{2}=11
25=11+2ab\therefore 25=11+2ab
ab=7\therefore ab=7,即abab的值为77
②令a=x2022a=x-2022
x2021\therefore x-2021
=[x(20221)]=\left[x-\left(2022-1\right)\right]
=x2022+1=x-2022+1
=a+1=a+1
x2023x-2023
=[x(2022+1)]=\left[x-\left(2022+1\right)\right]
=x20221=x-2022-1
=a1=a-1
(x2021)2+(x2023)2=20\because \left(x-2021\right)^{2}+\left(x-2023\right)^{2}=20
(a+1)2+(a1)2=20\therefore \left(a+1\right)^{2}+\left(a-1\right)^{2}=20
解得:a2=9a^{2}=9.
(x2022)2=9\therefore \left(x-2022\right)^{2}=9
x2022=±3\therefore x-2022=\pm 3.

解析

(1)大正方形的面积可以表示为:(a+b)2\left(a+b\right)^{2},或表示为:a2+b2+2aba^{2}+b^{2}+2ab
因此有(a+b)2=a2+b2+2ab\left(a+b\right)^{2}=a^{2}+b^{2}+2ab
(2)(2)根据题意得:(a+2b)(a+b)=a2+3ab+2b2\left(a+2b\right)\left(a+b\right)=a^{2}+3ab+2b^{2}
则需要AA号卡片11张,BB号卡片22张,CC号卡片33张.
故答案为:33
(3)(3)(a+b)2=a2+b2+2ab\because \left(a+b\right)^{2}=a^{2}+b^{2}+2aba+b=5a+b=5a2+b2=11a^{2}+b^{2}=11
25=11+2ab\therefore 25=11+2ab
ab=7\therefore ab=7,即abab的值为77
②令a=x2022a=x-2022
x2021\therefore x-2021
=[x(20221)]=\left[x-\left(2022-1\right)\right]
=x2022+1=x-2022+1
=a+1=a+1
x2023x-2023
=[x(2022+1)]=\left[x-\left(2022+1\right)\right]
=x20221=x-2022-1
=a1=a-1
(x2021)2+(x2023)2=20\because \left(x-2021\right)^{2}+\left(x-2023\right)^{2}=20
(a+1)2+(a1)2=20\therefore \left(a+1\right)^{2}+\left(a-1\right)^{2}=20
解得:a2=9a^{2}=9.
(x2022)2=9\therefore \left(x-2022\right)^{2}=9
x2022=±3\therefore x-2022=\pm 3.

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