题目内容
设3x3+(4-3
)x2-3
x-7=O,则x4+
x3-7x2-3
x+2的值为( )
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| 7 |
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A、30
| ||
| B、30 | ||
C、
| ||
| D、0 |
分析:由3x3+(4-3
)x2-3
x-7=O,可整理得,3x3-3
x2+4x2-4
x+
x-7=0,则3x2(x-
)+4x(x-
)+
(x-
)=0,所以,(x-
)(3x2+4x+
)=0,根据二次函数根的判别式可知,二次函数3x2+4x+
=0,无解,所以,x=
;代入、解答出即可;
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解答:解:由3x3+(4-3
)x2-3
x-7=O,
整理得,3x3-3
x2+4x2-4
x+
x-7=0,
3x2(x-
)+4x(x-
)+
(x-
)=0,
∴(x-
)(3x2+4x+
)=0,
∴x-
=0或3x2+4x+
=0,
∵二次函数3x2+4x+
=0,根的判别式△=42-4×3×
<0,
∴3x2+4x+
=0无解,
∴x=
,
∴x4+
x3-7x2-3
x+2=
4+
×
3-7×
2-3
×
+2=30;
故选B.
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整理得,3x3-3
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3x2(x-
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∴(x-
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∴x-
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∵二次函数3x2+4x+
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∴3x2+4x+
| 7 |
∴x=
| 7 |
∴x4+
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故选B.
点评:本题主要考查了高次方程的解法,对已知方程作适当分解、变形而求出未知数x,是解答出本题的关键.
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