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九年级数学解答题一般
题目
把代数式通过配方等手段得到完全平方式,再运用完全平方式的非负性这一性质解决问题,这种解题方法叫做配方法.配方法在代数式求值,解方程,最值问题等都有广泛的应用.如利用配方法求最小值,求a2+6a+8a^{2}+6a+8的最小值.
解:a2+6a+8=a2+6a+3232+8=(a+3)21a^{2}+6a+8=a^{2}+6a+3^{2}-3^{2}+8=\left(a+3\right)^{2}-1,因为不论aa取何值,(a+3)2\left(a+3\right)^{2}总是非负数,即(a+3)20\left(a+3\right)^{2}\geqslant 0.所以(a+3)211\left(a+3\right)^{2}-1\geqslant -1,所以当a=3a=-3时,a2+6a+8a^{2}+6a+8有最小值,最小值是1-1.
根据上述材料,解答下列问题:
(1)(1)填空:x210x+______=(x______)2x^{2}-10x+\_\_\_\_\_\_=(x-\_\_\_\_\_\_)^{2}
(2)(2)x28x+2x^{2}-8x+2变形为(x+m)2+n\left(x+m\right)^{2}+n的形式,并求出x28x+2x^{2}-8x+2的最小值;
(3)(3)M=4a2+9a+3M=4a^{2}+9a+3,N=3a2+11a1N=3a^{2}+11a-1,其中aa为任意数,试比较MMNN的大小,并说明理由.
知识点:配方法的应用、非负数的性质:偶次方章节:第2章 一元二次方程 / 2.2 用配方法求解一元二次方程

答案与解析

答案

(1)x210x+25=(x5)2\left(1\right)x^{2}-10x+25=\left(x-5\right)^{2}
故答案为:252555
(2)x28x+2(2)x^{2}-8x+2
=x28x+1616+2=x^{2}-8x+16-16+2
=(x4)214=\left(x-4\right)^{2}-14
\because不论xx取何值,(x4)2\left(x-4\right)^{2}总是非负数,
(x4)20\left(x-4\right)^{2}\geqslant 0
(x4)21414\therefore \left(x-4\right)^{2}-14\geqslant -14
\thereforex=4x=4时,x28x+2x^{2}-8x+2有最小值,最小值是14-14
(3)M>N(3)M \gt N.理由如下:
MNM-N
=4a2+9a+3(3a2+11a1)=4a^{2}+9a+3-(3a^{2}+11a-1)
=4a2+9a+33a211a+1=4a^{2}+9a+3-3a^{2}-11a+1
=a22a+4=a^{2}-2a+4
=a22a+11+4=a^{2}-2a+1-1+4
=(a1)2+3=\left(a-1\right)^{2}+3
(a1)20\because \left(a-1\right)^{2}\geqslant 0
(a1)2+3>0\therefore \left(a-1\right)^{2}+3 \gt 0
MN>0\therefore M-N \gt 0
M>N\therefore M \gt N.

解析

(1)x210x+25=(x5)2\left(1\right)x^{2}-10x+25=\left(x-5\right)^{2}
故答案为:252555
(2)x28x+2(2)x^{2}-8x+2
=x28x+1616+2=x^{2}-8x+16-16+2
=(x4)214=\left(x-4\right)^{2}-14
\because不论xx取何值,(x4)2\left(x-4\right)^{2}总是非负数,
(x4)20\left(x-4\right)^{2}\geqslant 0
(x4)21414\therefore \left(x-4\right)^{2}-14\geqslant -14
\thereforex=4x=4时,x28x+2x^{2}-8x+2有最小值,最小值是14-14
(3)M>N(3)M \gt N.理由如下:
MNM-N
=4a2+9a+3(3a2+11a1)=4a^{2}+9a+3-(3a^{2}+11a-1)
=4a2+9a+33a211a+1=4a^{2}+9a+3-3a^{2}-11a+1
=a22a+4=a^{2}-2a+4
=a22a+11+4=a^{2}-2a+1-1+4
=(a1)2+3=\left(a-1\right)^{2}+3
(a1)20\because \left(a-1\right)^{2}\geqslant 0
(a1)2+3>0\therefore \left(a-1\right)^{2}+3 \gt 0
MN>0\therefore M-N \gt 0
M>N\therefore M \gt N.

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