Question

已知x²=x+1，y²=y+1，且x≠y。求x的五次方加y的五次方的值。

Answer 1

已知x²=x+1，y²=y+1，且x≠y。求x的五次方加y的五次方的值。

x²=x+1，
x=(1±√5)/2

y²=y+1，
y=(1±√5)/2

x≠y
x+y=(1+√5)/2+(1-√5)/2=1

x^5+y^5
=x^2*x^2*x+y^2*y^2*y
=(x+1)(x+1)*x+(y+1)(y+1)*y
=x(x^2+2x+1)+y(y^2+2y+1)
=x(x+1+2x+1)+y(y+1+2y+1)
=x(3x+2)+y(3y+2)
=3x^2+2x+3y^2+2y
=3(x+1)+2x+3(y+1)+2y
=5x+3+5y+3
=5(x+y)+6
=5*1+6
=11