设x=(√5-1)/2,求x^4+x^2+2x-1的值.
1个回答

x=〔(√5)-1]/2

2x+1=√5

两边平方

4x^2+4x+1=5

4x^2+4x=4

x^2+x=1

x^4+x^2+2x-1

=x^4+x^3-x^3+x^2+2x-1

=x^2(x^2+x)-x^3+x^2+2x-1

=x^2*1-x^3+x^2+2x-1

=-x^3+2x^2+2x-1

=-x^3-x^2+3x^2+2x-1

=-x(x^2+x)+3x^2+2x-1

=-x*1+3x^2+2x-1

=3x^2+x-1

=3x^2+3x-2x-1

=3(x^2+x)-2x-1

=3*1-2x-1

=-2x+2

=-2*(√5-1)/2+2

=-√5+1+2

=3-√5