设x<0,且x-1⼀x=5 设x<0,且x-1⼀x=根号5,求代数式(x^10+x^6+x^4+1)⼀(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

请看图片：