设x<0,且x-1/x=5 设x<0,且x-1/x=根号5,求代数式(x^10+x^6+x^4+1)/(x^10+x^8+x^2+1)的值

x - 1/x = 根号5 ---> x^2 - 根号5*x -1 =0

解得x = (根号5 - 3)/2 (因为x小於0)

并令a = (根号5 - 3)/2

f(x) = x^2 - 根号5*x -1

b(x) = x^10+x^6+x^4+1

c(x) = x^10+x^8+x^2+1

则 f(a) = 0

利用长除法求得(可利用一点技巧)

b(x) / f(x)之余式为42( 55根号5 * x + 47) --->设商为g(x)

c(x) / f(x)之余式为47( 55根号5 * x + 47) --->设商为g'(x)

设原式 = F(x) = b(x)/c(x) = [ f(x) * g(x) + 42( 55根号5 * x + 47) ] / [ f(x) * g'(x) + 47( 55根号5 * x + 47) ]

则F(a) = [ f(a) * g(a) + 42( 55根号5 * a + 47) ] / [ f(a) * g'(a) + 47( 55根号5 * a + 47) ] = 42( 55根号5 * a + 47) / 47( 55根号5 * a + 47) = 42/47