己知方程x2-x-1=0的根是方程x6-px2+q=0的根,则p=________,q=________.
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己知方程x2-x-1=0的根是方程x6-px2+q=0的根,则p=________,q=________.
己知方程x2-x-1=0的根是方程x6-px2+q=0的根,则p=________,q=________.
己知方程x2-x-1=0的根是方程x6-px2+q=0的根,则p=________,q=________.
设方程x²-x-1=0的二根分别为x₁、x₂,由韦达定理,得:
x₁+x₂=1
x₁*x₂=-1
则:
x₁²+x₂²=(x₁+x₂)²-2x₁*x₂=1+2=3
(x₁²)²+(x₂²)²
=(x₁²+x₂²)²-2x₁²*x₂²
=(x₁²+x₂²)²-2(x₁*x₂)²
=3²-2*(-1)²
=7
将x₁、x₂分别代入方程x^6-px²+q=0得:
x₁^6-px₁²+q=0······①
x₂^6-px₂²+q=0······②
①-②,得:
(x₁^6-x₂^6)-p(x₁²-x₂²)=0
[(x₁²)³-(x₂²)³]-p(x₁²-x₂²)=0
(x₁²-x₂²)[(x₁²)²+(x₂²)²+x₁²*x₂²]-p(x₁²-x₂²)=0
由于x₁≠x₂,则x₁²-x₂²≠0,所以化简得:
[(x₁²)²+(x₂²)²+x₁²*x₂²]-p=0
则:
p=(x₁²)²+(x₂²)²+(x₁*x₂)²
=7+(-1)²
=8
①+②,得:
(x₁^6+x₂^6)-8(x₁²+x₂²)+2q=0
[(x₁²)³+(x₂²)³]-24+2q=0
(x₁²+x₂²)[(x₁²)²+(x₂²)²-x₁²*x₂²]-24+2q=0
3[(x₁²)²+(x₂²)²-(x₁*x₂)²]-24+2q=0
3(7-1)-24+2q=0
得:q=3;
综上,得:p=8,q=3.