题目内容
7、已知x2-2x-4=0,那么代数式x3+x2-10x+13的值为( )
分析:首先把已知条件x2-2x-4=0,可得到x2-2x=4,然后再把式子x3+x2-10x+13,进行变形,分解因式,逐步将x2-2x=4代入所变形的式子,即可得到答案.
解答:解:∵x2-2x-4=0,
∴x2-2x=4,
∴x3+x2-10x+13,
=x3-2x2+3x2-10x+13,
=x(x2-2x)+3x2-10x+13,
=4x+3x2-10x+13,
=3x2-6x+13,
=3(x2-2x)+13,
=3×4+13,
=25.
故选:B.
∴x2-2x=4,
∴x3+x2-10x+13,
=x3-2x2+3x2-10x+13,
=x(x2-2x)+3x2-10x+13,
=4x+3x2-10x+13,
=3x2-6x+13,
=3(x2-2x)+13,
=3×4+13,
=25.
故选:B.
点评:此题主要考查了因式分解的应用,又考查了代数式求值的方法,同时还隐含了整体的数学思想和正确运算的能力.
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