初三数学题？ 初三数学题？
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4. \u4e8c\u6b21\u51fd\u6570 \u7684\u6700\u5c0f\u503c\u662f\uff08 A \uff09
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6. \u4e0b\u5217\u8fd0\u7b97\u6b63\u786e\u7684\u662f\uff08 B \uff09
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11. \u5df2\u77e5\u51fd\u6570 \uff0c\u5f53 =1\u65f6\uff0c \u7684\u503c\u662f________2
12. \u5728\u67d0\u6821\u4e3e\u884c\u7684\u827a\u672f\u8282\u7684\u6587\u827a\u6f14\u51fa\u6bd4\u8d5b\u4e2d\uff0c\u4e5d\u4f4d\u8bc4\u59d4\u7ed9\u5176\u4e2d\u4e00\u4e2a\u8868\u6f14\u8282\u76ee\u73b0\u573a\u6253\u51fa\u7684\u5206\u6570\u5982\u4e0b\uff1a9.3\uff0c8.9\uff0c9.3\uff0c9.1\uff0c8.9\uff0c8.8\uff0c9.3\uff0c9.5\uff0c9.3\uff0c\u5219\u8fd9\u7ec4\u6570\u636e\u7684\u4f17\u6570\u662f________9.3
13. \u7edd\u5bf9\u503c\u662f6\u7684\u6570\u662f________+6,-6
14. \u5df2\u77e5\u547d\u9898\u201c\u5982\u679c\u4e00\u4e2a\u5e73\u884c\u56db\u8fb9\u5f62\u7684\u4e24\u6761\u5bf9\u89d2\u7ebf\u4e92\u76f8\u5782\u76f4\uff0c\u90a3\u4e48\u8fd9\u4e2a\u5e73\u884c\u56db\u8fb9\u5f62\u662f\u83f1\u5f62\u201d\uff0c\u5199\u51fa\u5b83\u7684\u9006\u547d\u9898\uff1a________________________________\u7565
15. \u5982\u56fe7-\u2460\uff0c\u56fe7-\u2461\uff0c\u56fe7-\u2462\uff0c\u56fe7-\u2463\uff0c\u2026\uff0c\u662f\u7528\u56f4\u68cb\u68cb\u5b50\u6309\u7167\u67d0\u79cd\u89c4\u5f8b\u6446\u6210\u7684\u4e00\u884c\u201c\u5e7f\u201d\u5b57\uff0c\u6309\u7167\u8fd9\u79cd\u89c4\u5f8b\uff0c\u7b2c5\u4e2a\u201c\u5e7f\u201d\u5b57\u4e2d\u7684\u68cb\u5b50\u4e2a\u6570\u662f________\uff0c\u7b2c \u4e2a\u201c\u5e7f\u201d\u5b57\u4e2d\u7684\u68cb\u5b50\u4e2a\u6570\u662f________2n+5
16. \u5982\u56fe8\u662f\u7531\u4e00\u4e9b\u76f8\u540c\u957f\u65b9\u4f53\u7684\u79ef\u6728\u5757\u642d\u6210\u7684\u51e0\u4f55\u4f53\u7684\u4e09\u89c6\u56fe\uff0c\u5219\u6b64\u51e0\u4f55\u4f53\u5171\u7531________\u5757\u957f\u65b9\u4f53\u7684\u79ef\u6728\u642d\u62104
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(3)\u8bbePE=x,PH=y,\u6613\u5f97BG=1-x,BF=1-y,FG=x+y-1,\u7531\u52fe\u80a1\u5b9a\u7406\uff0c\u5f97
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对称轴是-b/2a=1，所以b=-2a。
所以两根之和是-b/a=2。
已知A点横坐标-1，也就是其中一个根是-1，那么另一个根就是2-（-1）=3。
再利用两根之积是c/a=-3，所以c=-3a。
这样的话，a＞0，b和c都小于零，所以abc＞0，第一个正确。
第二个结论，可以吧b和c都代入，这样4a+2b+c=4a-4a-3a=-3a＜0。第二个错误。
第三个结论，依旧可以代入，代入化简得-16a平方＜16a，这个肯定正确，但我怀疑题是不是有问题，不等号左边是不是b平方-4ac。
第四个结论，应该是利用与y轴交点纵坐标在-2和-1之间，也就是c在-2和-1之间，把c换成-3a，可得结论正确。
第五个直接代换比较一下就可知道正确。
如图，1、3、4、5正确
1.
连接CD
则角BDC是直角坐标
所能△BCD是直角三角形
因为AC=BC
所以D是AB中点(等腰三角形的高和中线合一)
即AD=BD
连接OD
DO=BO
△BOD是等腰三角形,角DBO是和等腰△ABC的公共角,所以DO平行于AC
所以DO垂直于DF,即DF是切线
2.∠C=25°
则.∠COB=75°
AO=DO
.∠A=.∠ADO=1/2.∠COB=37.5°
因为长和宽之比为3：2.5=6：5，30枚邮票要以5：6的方式排列才能形成正方形。具体是排5列（纵列），和6排（横排），这样横着有3×5=15厘米，竖着有2.5×6=15
所以形成边长为15厘米的正方形
解：因为根号下的数是恒大于等于0的，所以x≥2011，所以|2010-x|=x-2010已知条件课化简为x-2010+√（x-2011）=x所以，√（x-2011）=2010，即x=2010²-2011所以x-2010²=2010²-2011-2010²=2011
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