求极限lim x→∞ ﹙3x-√ax²-x+1﹚=1/6,求a
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求极限limx→∞﹙3x-√ax²-x+1﹚=1/6,求a求极限limx→∞﹙3x-√ax²-x+1﹚=1/6,求a求极限limx→∞﹙3x-√ax²-x+1﹚=1/6,
求极限lim x→∞ ﹙3x-√ax²-x+1﹚=1/6,求a
求极限lim x→∞ ﹙3x-√ax²-x+1﹚=1/6,求a
求极限lim x→∞ ﹙3x-√ax²-x+1﹚=1/6,求a
a=9
3x-√ax²-x+1=(3x-√ax²-x+1)/1
分子分母同时乘以3x+√ax²-x+1得到(9x2-ax2+x-1)/3x+√ax²-x+1
分母项最高次是x,那么分母最高次也一定是x,否则极限为正无穷,那么9-a=0推出a=9
那个请问那个根号开到哪里?
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