问一个数学奥数a^2-3a-1=0,求 (a^2+1)(a^8+1) / (a^4+1)(a^6+1)的值.请高手解答,
2个回答
不管题有没有错,这个题是有解的.但奥数不是考计算量,所以一定有巧法,可能是系数或者符号抄错了吧.
a^2-3a-1=0得a^2=3a+1
a^4=(3a+1)^2=9a^2+6a+1=9(3a+1)+6a+1=33a+10
a^6=a^4*a^2=(33a+10)(3a+1)=99a^2+63a+10=99(3a+1)+63a+10=360a+109
a^8=a^6*a^2=(360a+109)(3a+1)=1080a^2+687a+109=1080(3a+1)+687a+109=3927a+1189
带入化简
(a^2+1)(a^8+1) / (a^4+1)(a^6+1)
=（3a+1)(3927a+1190)/(360a+110)(33a+11)
=(11781a^2+7497a+1190)/(11880a^2+7590a+1210)
=(35453a+11781+7497a+1190)/(35640a+11880+7590a+1210)
=(42950a+12971)/(43230a+13090)
最后解方程a^2-3a-1=0,把a的值带入上式求得.
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