Question

已知x平分-3x+1=0,求x平分+1/x平分+2=?,求2x^3-3x^2+2009的值!

Answer 1

x²-3x+1=0

等式两边同除以x

x- 3+1/x=0

x+ 1/x=3

x²+1/x²+2=(x+1/x)²=3²=9

x²-3x=-1

(x- 3/2)²=5/4

x=(3±√5)/2

2x³-3x²+2009

=2x³-6x²+2x+3x²-9x+3+7x+2006

=2x(x²-3x+1)+3(x²-3x+1)+7x+2006

=2x·0+3·0+7x+2006

=7x+2006

=7·(3±√5)/2 +2006

=(4033±7√5)/2

第二问化不掉x,只有求根了.