f(x)＝根号下ax�0�52+2x+1 值域为[0,+∽]所以要ax�0�52+2x+1≥0b�0�5-4ac≥04-4×a×-1≥0a≥-1所以它的定义域是R就这样~
2^2-4a<=0, a>=1定义域为Ｒ
a>0，且4-4a>=0所以0<=a<=1
