1. 已知a^2-a+1=0,则a^1995+1/a^1995=
2. 求证(b^2-3b-(b-1)√(b^2-4)+2)/(b^2+3b-(b+1)√(b^2-4)+2) * √((b+2)/(b-2))=(1-b)/(1+b)
3. 已知a+b=1. 求证： b/(a^3-1)-a/(b^3-1)=2(a-b)/(a^2b^2+3)

1. 已知a²-a+1=0,则a^1995+1/a^1995=

a²-a+1=0--->(a+1)(a²-a+1)=a³+1=0--->a³=-1
a^1995 = a^(3*665) = (a³)^665 = -1
--->a^1995+1/a^1995 = -2

2.
b²-3b-(b-1)√(b²-4)+2
=(b-1)(b-2)-(b-1)√(b²-4)
=(b-1)√(b-2)[√(b-2)-√(b+2)]

b²+3b-(b+1)√(b²-4)+2
=(b+1)(b+2)-(b+1)√(b²-4)
=(b+1)√(b+2)[√(b+2)-√(b-2)]

∴原题左边 = [-(b-1)√(b-2)]/[(b+1)√(b+2)]×√((b+2)/(b-2))=(1-b)/(1+b) = 右边

3. 已知a+b=1. 求证：b/(a³-1)-a/(b³-1)=2(a-b)/(a²b²+3)

　b/(a³-1) - a/(b³-1)
= (1-a)/(a³-1) - (1-b)/(b³-1)
= 1/(b²+b+1) - 1/(a²+a+1)
= [(a²+a+1)-(b²+b+1)]/[(b²+b+1)(a²+a+1)]
= [(a²-b²)+(a-b)]/(a²b²+ab²+a²b+b²+a²+ab+a+b+1)
= [(a-b)(a+b)+(a-b)]/[a²b²+ab(a+b)+(a+b)²-ab+(a+b)+1]
= 2(a-b)/(a²b²+3)