已知x2+4x-2=0则代数式x4+3x3-5x2+6x+15=

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已知x2+4x-2=0则代数式x4+3x3-5x2+6x+15=
已知x2+4x-2=0则代数式x4+3x3-5x2+6x+15=

已知x2+4x-2=0则代数式x4+3x3-5x2+6x+15=
x4+3x3-5x2+6x+15
=x²(x²+4x-2)-x³-3x²+6x+15
=-x³-3x²+6x+15
=-x(x²+4x-2)+x²+4x+15
=x²+4x+15
=x²+4x-2+17
=17

解
x²+4x-2=0
x²+4x=2
∴x^4+3x³-5x²+6x+15
=x²(x²+4x)-x³-5x²+6x+15
=2x²-x³-5x²+6x+15
=-x³-3x²+6x+15
=-x(x²+4x)+x²+6x+15
=-2x+x²+6x+15
=x²+4x+15
=2+15
=17望采纳

x^2+4x=2
x^4+3x^3-5x^2+6x+15
=x^2(x^2+4x)-4x^3+3x^3-5x^2+6x+15
=2x^2-x^3-5x^2+6x+15
=-x(x^2+4x)+4x^2-3x^2+6x+15
=-2x+x^2+6x+15
=x^2+4x+15
=2+15
=17

x⁴+3x³-5x²+6x+15
=x⁴+4x³-2x²-x³-4x²+2x+x²+4x-2+17
=x²(x²+4x-2)-x(x²+4x-2)+(x²+4x-2)+17
=x²·0-x·0+0+17
=17

x4+3x3-5x2+6x+15
=x^4+4x³-2x² -x³-4x²+2x +x²+4x-2+17
=x²(x²+4x-2)-x(x²+4x-2)+(x²+4x-2)+17
∵x²+4x-2=0
∴原式=x²×0-x×0+0+17=17

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