试题

题目:
已知x=4-
3
,求
x4-6x3-2x2+18x+23
x2-8x+15
的值.
答案
解:已知得(x-4)2=3,即x2-8x+13=0,则x2-8x=-13.
分子x4-6x3-2x2+18x+23,
=x4-8x3+2x3-2x2+18x+23,
=x2(x2-8x)+2x3-2x2+18x+23,
=-13x2+2x3-2x2+18x+23,
=2x3-16x2+x2+18x+23,
=2x(x2-8x)+x2+18x+23,
=-26x+x2+18x+23,
=x2-8x+23,
=-13+23,
=10,
分母是x2-8x+15=-13+15=2,
x4-6x3-2x2+18x+23
x2-8x+15
=
10
2
=5.
故答案为:5.
解:已知得(x-4)2=3,即x2-8x+13=0,则x2-8x=-13.
分子x4-6x3-2x2+18x+23,
=x4-8x3+2x3-2x2+18x+23,
=x2(x2-8x)+2x3-2x2+18x+23,
=-13x2+2x3-2x2+18x+23,
=2x3-16x2+x2+18x+23,
=2x(x2-8x)+x2+18x+23,
=-26x+x2+18x+23,
=x2-8x+23,
=-13+23,
=10,
分母是x2-8x+15=-13+15=2,
x4-6x3-2x2+18x+23
x2-8x+15
=
10
2
=5.
故答案为:5.
考点梳理
分式的化简求值.
由已知得(x-4)2=3,即x2-8x+13=0,则x2-8x=-13,把分子、分母变形利用x2-8x表示,代入求值即可.
本题使用了整体代换的方法.正确对分子进行变换是解题的关键.
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