(1)因?yàn)槠婧瘮?shù)（ax^2+bx+1）/（cx+d）為奇函數(shù),所以f(-x)+f(x)=0恒成立可得b=d=0,
所以f(x)=(ax^2+1)/cx,f(1)=(a+1)/c=3,得a+1=3c①,又當(dāng)x大于0,f(x)的最小值為2根號(hào)2,
f(x)=(ax^2+1)/cx=a(x+1/(ax))/c,所以c>0,由均值不等式可知當(dāng)x=1/√a時(shí)等號(hào)成立,
所以2√a/c=2√2,得a=2c^2②,聯(lián)立①②可求c=1/2(此時(shí)a=1/2,不合舍去)或c=1,a=2
所以f(x)=(2x^2+1)/x
(2) a(n+1)的平方/an=f(an)=(2an^2+1)/an→a(n+1)^2=2a(n)^2+1,
所以a(n+1)^2+1=2[a(n)^2+1],所以數(shù)列｛a(n)^2+1｝是首項(xiàng)等于6,公比等于2的等比數(shù)列,
a(n)^2+1=3×2^n,
所以an=√(3×2^n-1)
(1)bn=n,數(shù)列{bncosnπ}的前n項(xiàng)和Tn=-1+2-3+4+----+ncosnπ=①當(dāng)n=4k時(shí),Tn=2k；②當(dāng)n=4k+1時(shí),Tn=2k-1；③當(dāng)n=4k+2時(shí),Tn=2k+1；④當(dāng)n=4k+3時(shí),Tn=2k-2