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七年级数学填空题一般
题目
整体代换是数学的一种思想方法,例如:x2+x=0x^{2}+x=0,则x2+x+1186=x^{2}+x+1186=______;
我们将x2+xx^{2}+x作为一个整体代入,则原式=0+1186=1186=0+1186=1186.
仿照上面的解题方法,完成下面的问题:
(1)(1)x2+x3=0x^{2}+x-3=0,则x2+x+2021=x^{2}+x+2021=______;
(2)(2)如果a+b=6a+b=6,求2(a+b)4a4b+212\left(a+b\right)-4a-4b+21的值;
(3)(3)a2+2ab=22a^{2}+2ab=22,b2+2ab=8b^{2}+2ab=8,求2a23b22ab2a^{2}-3b^{2}-2ab的值.
知识点:换元法解分式方程章节:未标注

答案与解析

答案

由题可知,:x2+x=0x^{2}+x=0,则x2+x+1186=1186x^{2}+x+1186=1186
(1)x2+x3=0(1)\because x^{2}+x-3=0
x2+x=3\therefore x^{2}+x=3
x2+x+2021=3+2021=2024\therefore x^{2}+x+2021=3+2021=2024.
故答案为:20242024
(2)a+b=6(2)\because a+b=6
2(a+b)4a4b+21\therefore 2\left(a+b\right)-4a-4b+21
=2(a+b)4(a+b)+21=2\left(a+b\right)-4\left(a+b\right)+21
=2(a+b)+21=-2\left(a+b\right)+21
=2×6+21=-2\times 6+21
=12+21=-12+21
=9=9
(3)a2+2ab=22(3)\because a^{2}+2ab=22b2+2ab=8b^{2}+2ab=8
a2=222ab\therefore a^{2}=22-2abb2=82abb^{2}=8-2ab
2a23b22ab\therefore 2a^{2}-3b^{2}-2ab
=2(222ab)3(82ab)2ab=2\left(22-2ab\right)-3\left(8-2ab\right)-2ab
=444ab24+6ab2ab=44-4ab-24+6ab-2ab
=20=20.
故答案为:11861186(1)2024\left(1\right)2024(2)9\left(2\right)9(3)20\left(3\right)20.

解析

由题可知,:x2+x=0x^{2}+x=0,则x2+x+1186=1186x^{2}+x+1186=1186
(1)x2+x3=0(1)\because x^{2}+x-3=0
x2+x=3\therefore x^{2}+x=3
x2+x+2021=3+2021=2024\therefore x^{2}+x+2021=3+2021=2024.
故答案为:20242024
(2)a+b=6(2)\because a+b=6
2(a+b)4a4b+21\therefore 2\left(a+b\right)-4a-4b+21
=2(a+b)4(a+b)+21=2\left(a+b\right)-4\left(a+b\right)+21
=2(a+b)+21=-2\left(a+b\right)+21
=2×6+21=-2\times 6+21
=12+21=-12+21
=9=9
(3)a2+2ab=22(3)\because a^{2}+2ab=22b2+2ab=8b^{2}+2ab=8
a2=222ab\therefore a^{2}=22-2abb2=82abb^{2}=8-2ab
2a23b22ab\therefore 2a^{2}-3b^{2}-2ab
=2(222ab)3(82ab)2ab=2\left(22-2ab\right)-3\left(8-2ab\right)-2ab
=444ab24+6ab2ab=44-4ab-24+6ab-2ab
=20=20.
故答案为:11861186(1)2024\left(1\right)2024(2)9\left(2\right)9(3)20\left(3\right)20.

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