Question

已知m是方程x2-3x+1=0的根,求下列代数式的值(1)m3-8m+1

已知m是方程x2-3x+1=0的根,求下列代数式的值(1)m3-8m+1 （2)m2+m2分之1 (3)m分之1-m-3分之1（4）m2-2m+(m2+1 分之3)

要详细过程，速度一点点，谢谢，好的有悬赏

Answer 1

解：
x=m代入方程
m²-3m+1=0
m²=3m-1
(1)
m³-8m+1
=m×m²-8m+1
=m×(3m-1)-8m+1
=3m²-m-8m+1
=3(3m-1)-9m+1
=9m-3-9m+1
=-2
(2)
m² -3m+1=0
等式两边同除以m
m -3 +1/m=0
m +1/m=3
m²+1/m²=(m+1/m)²-2=3²-2=9-2=7
(3)
m²-3m+1=0
m²-3m=-1
1/m -1/(m-3)
=[(m-3)-m]/[m(m-3)]
=(m-3-m)/(m²-3m)
=(-3)/(-1)
=3
(4)
m²-3m+1=0
m²+1=3m
m²-2m=m-1
m²-2m+ 3/(m²+1)
=m-1 +3/(3m)
=m +1/m -1
=3 -1
=2

Answer 2

m是方程x^2-3x+1=0的根
m^2-3m+1=0
m^2-3m=-1
m^2=3m-1

m^3-8m+1
=m*m^2-8m+1
=m(3m-1)-8m+1
=3m^2-m-8m+1
=3m^2-9m+1
=3(m^2-3m)+1
=3*(-1)+1
=-3+1
=-2

m^2-3m+1=0
m-3+1/m=0
m+1/m=3 平方
(m+1/m)^2=9
m^2+1/m^2+2=9
m^2+1/m^2=7

1/m-1/(m-3)
=(m-3)/[m(m-3)]-m/[m(m-3)]
=(m-3-m)/[m(m-3)]
=-3/(m^2-3m)
=-3/(-1)
=3

m^2-3m+1=0
m^2+1=3m

m^2-2m+3/(m^2+1)
=m^2-2m+3/(3m)
=m^2-2m+1/m
=m^2-3m+m+1/m
=-1+3
=2