已知x^2+x-1=0,求2x^4+3x^2+2⼀x^3+2x^2-x的值.

已知x^2+x-1=0,求2x^4+3x^2+2/x^3+2x^2-x的值
2024-12-14 23:24:00
推荐回答(1个)
回答1:

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