證明：方程ax²-ax+1=0的判別式△=（-a)²-4a×1=a(a-4)
當(dāng)a＜0時(shí),△＞0,函數(shù)y=ax²-ax+1開(kāi)口向下,與x軸有兩個(gè)交點(diǎn),
∴y=ax²-ax+1可正可負(fù),對(duì)于一切實(shí)數(shù)X不一定都成立.
當(dāng)0＜a＜4時(shí),△＜0,函數(shù)y=ax²-ax+1開(kāi)口向上,與x軸沒(méi)有交點(diǎn),
∴y=ax²-ax+1＞0,即對(duì)于一切實(shí)數(shù)X,ax²-ax+1＞0一定都成立.
當(dāng)a＞4時(shí),△＞0,函數(shù)y=ax²-ax+1開(kāi)口向上,與x軸有兩個(gè)交點(diǎn),
∴y=ax²-ax+1可正可負(fù),對(duì)于一切實(shí)數(shù)X不一定都成立.
所以,關(guān)于X的一元二次不等式ax²-ax+1＞0,對(duì)于一切實(shí)數(shù)X都成立的重要條件是
0＜a＜4