Question

设x1,x2是方程3x2-4x=-1的两根,不解方程求下列各式的值 (1) ∣x1-x2∣ (2)9x13+13

Answer 1

x1,x2是方程3x^2-4x+1=0的两根，
所以x1+x2=4/3,x1x2=1/3.
(1)|x1-x2|=√(x1-x2)^2
=√[(x1+x2)^2-4x1x2]
=√(16/9-4/3)
=√(4/9)
=2/3.
(2)x^2=(4x-1)/3,
x^4=[(4x-1)/3]^2
=(16x^2-8x+1)/9
=(64x-16-24x+3)/27
=(40x-13)/27,
x^8=[(40x-13)/27]^2
=(1600x^2-1040x+169)/729
=(6400x-1600-3120x+507)/2187
=(3280x-1093)/2187,
x^13=x^8*x^4*x
=(3280x-1093)(40x-13)x/59049
=(3280x-1093)(121x-40)/177147
=(396880x^2-263453x+43720)/177147
=(1587520x-396880-790359x+131160)/531441
=(797161x-265720)/531441,
9x^13+13=(797161x-265720)/59049+13,
需解方程，题目欠完整。

Answer 2

x1,x2是方程3x2-4x=-1的两根，所以X1+X2=4/3,X1X2=1/3,
(X1+X2)^2=16/9,
(X1-X2)^2=16/9-4X1X2=16/9-4/3=4/9
X1-X2=2/3