已知x的平方-2x-1=0,求x4+x3-5x2-7x+5

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已知x的平方-2x-1=0,求x4+x3-5x2-7x+5

已知x的平方-2x-1=0,求x4+x3-5x2-7x+5
已知x的平方-2x-1=0,求x4+x3-5x2-7x+5

已知x的平方-2x-1=0,求x4+x3-5x2-7x+5
x^2-2x-1=0
x^2-2x=1
x^4+x^3-5x^2-7x+5
=x^2(x^2-2x)+3x^3-5x^2-7x+5
=x^2+3x^3-5x^2-7x+5
=3x^3-4x^2-7x+5
=3x^3-6x^2+2x^2-7x+5
=3x(x^2-2x)+2x^2-7x+5
=3x+2x^2-7x+5
=2x^2-4x+5
=2(x^2-2x)+5
=2+5
=7

x的平方-2x-1=0,求x4+x3-5x2-7x+5
x^4-2x^3-x^2=0
3x^3-6x^2-3x=0
2x^-4x-2=0
三式相加
x^4+x^3-5x^2-7x-2=0
所以x4+x3-5x2-7x+5=7

因为x2-2x-1=0
所以x2-x=x+1
所以原式=x4-x3+2x3-2x2-(3x2-3x)-10x+5
=x2(x+1)+2x(x+1)-3(x+1)+5-10x
=x3+3x2-11x+2
=x3-x2+4x2-4x-7x+2
=x(x+1)+4(x+1)-7x+2
=x2-2x+6
=x2-x+6-x
=x+1+6-x
=7