作业帮2x=1-√5,求x^30-2x^28-x^27
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2x=1-√5,求x^30-2x^28-x^27
2x=1-√5,求x^30-2x^28-x^27
2x=1-√5,求x^30-2x^28-x^27
由题得 x^30-2x^28-x^27=x^27(x^3-2x-1)
=x^27(x^3-x^2-X+1-2-X+x^2)
=x^27[(x^3-x^2-X+1)+(x^2--X-2)]
=x^27[(X-1)^2(X+1)+(X+1)(X-2)]
=x^27[(X+1)(X^2-2x+1+X-2)]
=x^27[(X+1)(X^2-x-1)]
=x^27[X^2(X+1)-(x+1)^2]
X=(1-√5)/2 x^2==(1-√5)^2/4=(3--√5)/2
X+1=(3--√5)/2 (x+1)^2=(3-√5)^2/4
x^30-2x^28-x^27 =x^27[X^2(X+1)-(x+1)^2]
=x^27【(3--√5)/2(3--√5)/2 - (3-√5)^2/4】
=0
x=(1-根号5)/2
X^2=(1-2根号5+5）/4=(3-根号5）/2
所以, ,X^2-X=1,X^2-X-1=0, X^2-1=X
那么,X^30-2X^28-X^27
=(X^30-X^28)-(X^28+X^27)
=X^28*(X^2-1)-(X^28+X^27)
=X^28*X-X^28-X^27
=X^27(X^2-1)-X^28
=X^27*X-X^28
=X^28-X^28
=0
化简欲求多项式为x^26（x^4－2*x^2－x） 因为x^2=（3－√5）÷2；x^4=（7－3*√5）÷2；所以x^4－2*x^2－x＝0 所以欲求式为0