奥数题（分式）十万火急！！！！！！！！！！！！！！！！！

上自习去了 回来解答 O(∩_∩)O~
1>令a=tanA b=tanB c=tanC d=tanD，由已知
tan(A-B)+tan(C-D)=0
=>tan（A-B)=-tan（C-D)
=>A-B=D-C
=>A-D=B-C
=>tan(A-D)=tan(B-C)
=>a-d/(1+ad)+(c-b)/(1+cd)=0
同理A+C=B+D
=>tan(A+C)=tan(B+D)
=>a+c/(1-ac)+(b+d)/(bd-1)=0
2> 依题意x是不能＝0的
由x/(x^2-ax+1)=1得到
1/(x+1/x)-a=1
=>x+1/x=a+1
=>(x+1/x)^2=x^2+1/x^2+2=(a+1)^2
然后用(x^2+1/x^2)(x+1/x)=x^3+1/x^3+x+1/x=x^3+1/x^3+a+1=[(a+1)^2-2]*(a+1)
=>x^3+1/x^3=(a+1)*(a^2+2a-2)=a^3+3a^2-2
又因为x^3/(x^6-a^3x^3+1)=1/(x^3+1/x^3-a^3)
=> 1/(x^3+1/x^3-a^3)=1/(a^3+3a^2-2-a^3)=1/(3a^2-2)
所以x^3/(x^6-a^3x^3+1)的值为 1/(3a^2-2)
3>由条件知：(b^2+c^2-a)/2bc<=1
=>(b-c)^2<=a^2
=>|b-c|<=a
同理|c-a|<=b |a-b|<=c
所以可设a b c 为三角形的三条边，则
(b^2+c^2-a^2)/2bc+(c^2+a^2-b^2)/2ac+(a^2+b^2-c^2/2ab=1 等同于
cosA+cosB+cosC=1
又A+B+C=180
=>cosB+cosC-cos(B+C)=1
=>2cos(B+C)/2-