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七年级数学填空题一般
题目
理解与思考:整体代换是数学的一种思想方法.例如:若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.仿照这样的解题方法,完成下面的问题:
(1)(1)x2+x1=0x^{2}+x-1=0,则x2+x+2024=x^{2}+x+2024=______;
(2)(2)a+b=3a+b=3,求2(a+b)ab+212\left(a+b\right)-a-b+21的值;
(3)(3)a2+2ab=20a^{2}+2ab=20,b2+2ab=8b^{2}+2ab=8,则a2+b2+4ab=a^{2}+b^{2}+4ab=______;
(4)(4)x=1x=1时,代数式ax5+bx3+cx5ax^{5}+bx^{3}+cx-5的值为mm,求当x=1x=-1时,代数式ax5+bx3+cx5ax^{5}+bx^{3}+cx-5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

(1)x2+x1=0\left(1\right)\because x^{2}+x-1=0
x2+x=1\therefore x^{2}+x=1
x2+x+2024=1+2024=2025\therefore x^{2}+x+2024=1+2024=2025
故答案为:20252025
(2)a+b=3(2)\because a+b=3
2(a+b)ab+21\therefore 2\left(a+b\right)-a-b+21
=2a+2bab+21=2a+2b-a-b+21
=a+b+21=a+b+21
=3+21=3+21
=24=24
(3)a2+2ab=20(3)\because a^{2}+2ab=20b2+2ab=8b^{2}+2ab=8
a2+b2+4ab\therefore a^{2}+b^{2}+4ab
=a2+2ab+b2+2ab=a^{2}+2ab+b^{2}+2ab
=20+8=20+8
=28=28
故答案为:2828
(4)(4)\becausex=1x=1时,代数式ax5+bx3+cx5ax^{5}+bx^{3}+cx-5的值为mm
a+b+c5=m\therefore a+b+c-5=m
a+b+c=m+5\therefore a+b+c=m+5
x=1x=-1时,
ax5+bx3+cx5ax^{5}+bx^{3}+cx-5
=abc5=-a-b-c-5
=(a+b+c)5=-\left(a+b+c\right)-5
=(m+5)5=-\left(m+5\right)-5
=m55=-m-5-5
=m10=-m-10.

解析

(1)x2+x1=0\left(1\right)\because x^{2}+x-1=0
x2+x=1\therefore x^{2}+x=1
x2+x+2024=1+2024=2025\therefore x^{2}+x+2024=1+2024=2025
故答案为:20252025
(2)a+b=3(2)\because a+b=3
2(a+b)ab+21\therefore 2\left(a+b\right)-a-b+21
=2a+2bab+21=2a+2b-a-b+21
=a+b+21=a+b+21
=3+21=3+21
=24=24
(3)a2+2ab=20(3)\because a^{2}+2ab=20b2+2ab=8b^{2}+2ab=8
a2+b2+4ab\therefore a^{2}+b^{2}+4ab
=a2+2ab+b2+2ab=a^{2}+2ab+b^{2}+2ab
=20+8=20+8
=28=28
故答案为:2828
(4)(4)\becausex=1x=1时,代数式ax5+bx3+cx5ax^{5}+bx^{3}+cx-5的值为mm
a+b+c5=m\therefore a+b+c-5=m
a+b+c=m+5\therefore a+b+c=m+5
x=1x=-1时,
ax5+bx3+cx5ax^{5}+bx^{3}+cx-5
=abc5=-a-b-c-5
=(a+b+c)5=-\left(a+b+c\right)-5
=(m+5)5=-\left(m+5\right)-5
=m55=-m-5-5
=m10=-m-10.

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