已知x^2+x-1=0,求2x^4+3x^2+2/x^3+2x^2-x的值.
答案:2 悬赏:60
解决时间 2021-02-15 08:52
- 提问者网友:℡她的他i☆
- 2021-02-14 15:51
已知x^2+x-1=0,求2x^4+3x^2+2/x^3+2x^2-x的值.
最佳答案
- 二级知识专家网友:野性且迷人
- 2021-02-14 16:18
x^2+x-1=0 x^2=1-x............1
2x^4+3x^2+2/x^3+2x^2-x
分子:2x^4+2x^2+x^2+2
=2x^2(x^2+1)+x^2+2将1式代入:
=2(1-x)(1-x+1)+1-x+2
=2(1-x)(2-x)+3-x
=2(x^2-3x+2)+3-x
=2(1-x-3x+2)+3-x
=2(-4x+3)+3-x
=-8x+6+3-x
=-9x+9
分母:x^3+2x^2-x
=x(x^2+2x-1)
=x(1-x+2x-1)
=x^2
=1-x
原式=(-9x+9)/(1-x)
=9(1-x)/(1-x)
=9
2x^4+3x^2+2/x^3+2x^2-x
分子:2x^4+2x^2+x^2+2
=2x^2(x^2+1)+x^2+2将1式代入:
=2(1-x)(1-x+1)+1-x+2
=2(1-x)(2-x)+3-x
=2(x^2-3x+2)+3-x
=2(1-x-3x+2)+3-x
=2(-4x+3)+3-x
=-8x+6+3-x
=-9x+9
分母:x^3+2x^2-x
=x(x^2+2x-1)
=x(1-x+2x-1)
=x^2
=1-x
原式=(-9x+9)/(1-x)
=9(1-x)/(1-x)
=9
全部回答
- 1楼网友:初心未变
- 2021-02-14 16:41
x^2+x-1=0
则 x^2+x=1
x^3+2x^2+3
=x*(x^2+x)+x^2+3
=x+x^2+3
=1+3
=4
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