若[x-(√a)/x²]^6展开式的常数项为60,则a等于T‹r+1›=C(6,r)[x^(6-r)][(√a)/x²]^r=C(6,r)[(√a)^r][x^(6-r-2r)=C(6,r)[(√a)^r][x^(6-3r)=60；6-3r=0,故r=2,于是有C(6,2)a=60,故a=60/C(6,2)=6...