数学题目:已知三角形ABC的两边是关于x的方程x²－3mx＋9m＝0的两根，第三边长是4，当m为何
\u5df2\u77e5\u25b3ABC\u7684\u4e24\u8fb9\u662f\u5173\u4e8ex\u7684\u65b9\u7a0bx²-3mx+9m=0\u7684\u4e24\u6839\uff0c\u7b2c\u4e09\u8fb9\u957f\u662f4\uff0c\u5f53m\u4e3a\u4f55\u503c\u65f6\uff0c\u25b3ABC\u662f\u7b49\u8170\u4e09\u89d2\u5f62\uff1f\u82e5\u7ed3\u8bba\u6210\u7acb\uff0c\u5219\u65b9\u7a0b\u7684\u4e24\u4e2a\u5b9e\u6570\u6839\u6709\u4e24\u79cd\u60c5\u51b5\uff1a
\u4e00\uff1a\u5176\u4e2d\u4e00\u4e2a\u5b9e\u6570\u6839\u4e3a4,\uff1a
\u5373\u539f\u65b9\u7a0b\u53ef\u5206\u89e3\u4e3a\uff08x-4\uff09\u00d7\uff08x-n\uff09=0 \u5373 x²-\uff084+n\uff09x+4n=0
\u53ef\u77e54+n=3m\uff0c 4n=9m\uff0c \u89e3\u5f97m=16/3
\u4e8c\uff1a\u6709\u4e24\u4e2a\u76f8\u7b49\u7684\u5b9e\u6570\u6839\uff1a\u5373\u539f\u65b9\u7a0b\u53ef\u53d8\u5f62\u4e3a\uff08x-n)²=0 \u5373 x²-2nx+n²=0
\u53ef\u77e52n=3m\uff0cn²=9m \u89e3\u5f97m=4 \u548cm=0\uff08m=0\u4e0d\u5408\u9898\u610f\uff0c\u820d\u53bb\uff09
\u25b3=9m²-36m\u22650
\u89e3\u5f97m\u22640\u6216\u22654
\u82e5\u4e00\u6839\u8fb9\u4e3a4\uff0c\u5e26\u5165\u65b9\u7a0bm=16/3\uff0c\u5e76\u89e3\u5f97\u53e6\u4e00\u6839\u4e3a12\u3002\u4e0d\u80fd\u6784\u6210\u4e09\u89d2\u5f62\u3002\u820d\u53bb
\u82e5\u4e24\u6839\u76f8\u7b49\uff0c\u25b3=0\uff0c\u5219m=0\u62164.
\u82e5m=0\uff0c\u5219x1=x2=0\uff0c\u820d\u53bb\u3002
\u82e5m=4\uff0c\u5219x1=x2=6.\u80fd\u6784\u6210\u4e09\u89d2\u5f62\u3002\u6240\u4ee5m=4
\u6216\u8005
\u8bbe\u4e24\u8fb9\u5206\u522b\u4e3ax1,x2
\u6240\u4ee5x1x2=9m
x1+x2=3m
\u8bbex1=x2
\u6240\u4ee5 2x1=3m , x1^2=9m
\u89e3\u7684x1=x2=6
\u6240\u4ee512=3m,m=4\uff08\u4e09\u8fb9\u662f6.6.4\u6ee1\u8db3\u4e24\u8fb9\u4e4b\u548c\u5927\u4e8e\u7b2c\u4e09\u8fb9\uff09
\u8bbex1\u4e3a4
\u52194+x2=3m
4x2=9m
\u89e3\u7684m=-3/16(x1,x2\u4e3a\u8d1f\u6570)
\u6240\u4ee5m=4
第二种情况：关于x的方程是两个不等实根，则其中一个实根一定等于4；则将x=4带入原方程，解出m=16/3；自己还要带进去看是否成立
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