初三数学题
\u521d\u4e09\u6570\u5b66\u9898\uff01\u4f60\u597d,\u5f88\u9ad8\u5174\u66ff\u4f60\u89e3\u7b54\u8fd9\u4e2a\u95ee\u9898\uff0c\u9996\u5148\uff0c\u8bbe\u8fd9\u4e2a\u56db\u8fb9\u5f62\u7684\u5bf9\u89d2\u7ebf\u957f\u5206\u522b\u4e3aa\u548cb\uff0c\u7531\u4e8e\u5bf9\u89d2\u7ebf\u76f8\u4e92\u5782\u76f4\u4e14\u9762\u79ef\u662f3600\uff0c\u6240\u4ee5\u5f97\u52301/2(ab)=3600,\u5373ab=7200\u3002\u800c\u7531a+b\u22652\u6839\u53f7ab,\u5219\u7ecf\u8fc7\u8f6c\u5316\u5f97\u5230\uff0cab\u22641/4(a+b)²,\u53737200*4\u2264\uff08a+b)²,\u7136\u540e\u4e24\u8fb9\u540c\u65f6\u5f00\u6839\u53f7\u5f97a+b\u2265120\u6839\u53f72\uff0c\u6240\u4ee5\u7ed3\u679cx=120\u6839\u53f72\u3002\u7075\u6d3b\u8fd0\u7528\u5c31\u597d\u554a~~~~~
\u89e3\uff1a\uff081\uff09\u5f53\u70b9P\u8fd0\u52a8\u5230\u2220ABC\u5f97\u5e73\u5206\u7ebf\u4e0a\u65f6\uff0c\u8fde\u63a5DP\uff0c\u6c42DP\u7684\u957f\u3002
\u6c42DP \u89e3\u6cd5\u4e00\uff1a
\u7531\u9898\u610f\uff0c\u5728 Rt\u25b3ABC \u4e2d\uff0c
\u2220ABC = 60\u00b0 \uff0cAB = 2\u221a3\uff0c
\u7531 sin\u2220ABC = AC / AB \u5f97\uff1a
AC = AB \u00d7 sin\u2220ABC
= 2\u221a3 \u00d7 sin60\u00b0
= 2\u221a3 \u00d7 \uff08\u221a3/2\uff09
= 3
\u7531 cos\u2220ABC = BC / AB \u5f97\uff1a
BC = AB \u00d7 cos\u2220ABC
= AB \u00d7 cos60\u00b0
= 2\u221a3 \u00d7 \uff081/2\uff09
= \u221a3
\u2235 BP \u5e73\u5206 \u2220ABC\uff0c
\u2234 \u2220PBC = \uff081/2\uff09\u00d7 \u2220ABC
= \uff081/2\uff09\u00d7 60\u00b0
= 30\u00b0
\u5728 Rt\u25b3PBC \u4e2d\uff0c
PC = BC \u00d7 tan\u2220PBC
= BC \u00d7 tan30\u00b0
= \u221a3 \u00d7 (\u221a3/3)
= 1
\u5728\u7b49\u8170\u76f4\u89d2\u4e09\u89d2\u5f62ADC\u4e2d\uff0c
\u8fc7\u70b9D \u4f5cDE \u22a5 AC \u4e0e \u70b9E\uff0c
\u5219\uff1aDE = EC = (1/2) \u00d7 AC = (1/2) \u00d7 3 = 3/2
\u2234 EP = EC -- PC
= 3/2 -- 1
= 1/2
\u5728Rt\u25b3DEP \u4e2d\uff0c\u7531\u52fe\u80a1\u5b9a\u7406\u5f97\uff1a
DP\u65b9 = DE\u65b9 + EP\u65b9
= (3/2)\u65b9 + (1/2)\u65b9
= 10 / 4
\u2234 DP = \u221a(10/4) = (\u221a10) / 2
\u4ee5\u4e0a\u89e3\u7b54\u4e2d\uff0c\u60a8\u4e5f\u53ef\u4ee5\u7531\u201c\u5728\u76f4\u89d2\u4e09\u89d2\u5f62\u4e2d\uff0c30\u00b0\u6240\u5bf9\u7684\u76f4\u89d2\u8fb9\u7b49\u4e8e\u659c\u8fb9\u7684\u4e00\u534a\u201d\u76f4\u63a5\u5f97\u51faBC = AB/2 = \u221a3\u3002\u8fdb\u800c\u7528\u52fe\u80a1\u5b9a\u7406\u6c42\u51faAC=3\u3002
\u6c42DP \u89e3\u6cd5\u4e8c\uff1a\u9002\u7528\u9ad8\u4e2d\u77e5\u8bc6\u201c\u4f59\u5f26\u5b9a\u7406\u201d\u3002
\u5728\u7b49\u8170\u76f4\u89d2\u25b3ADC\u4e2d\uff0cDC = AC \u00d7 cos\u2220DCA
= AC \u00d7 cos45\u00b0
= 3 \u00d7 (\u221a2/2)
= (3\u221a2) / 2
\u2234 DC\u65b9 = [ (3\u221a2) / 2 ]\u65b9 = 9/2
\u2234 DP\u65b9 = DC\u65b9 + PC\u65b9 -- 2 \u00d7 DC \u00d7 PC \u00d7 cos\u2220DCA
= 9/2 + 1 -- 2 \u00d7 [ (3\u221a2) / 2 ] \u00d7 1 \u00d7 cos45\u00b0
= 9/2 + 1 -- 2 \u00d7 [ (3\u221a2) / 2 ] \u00d7 1 \u00d7 (\u221a2/2)
= 9/2 + 1 -- 3
= 5/2
\u2234 DP = \u221a(5/2) = (\u221a10) / 2\u3002
\uff082\uff09\u5f53\u70b9P\u5728\u8fd0\u52a8\u8fc7\u7a0b\u4e2d\u51fa\u73b0DP=BC\u65f6\uff0c
\u6b64\u65f6\u2220PDA\u7684\u5ea6\u6570\u4e3a\uff1a15\u00b0 \u6216 75\u00b0 \uff0c\u9700\u5206\u522b\u8ba8\u8bba\uff1a
\u5728\u7b49\u8170\u76f4\u89d2\u4e09\u89d2\u5f62ADC\u4e2d\uff0c\u2220DAP = 45\u00b0
\u8fc7\u70b9D \u4f5cDE \u22a5 AC \u4e0e \u70b9E\uff0c
\u5219\uff1aDE = EC = (1/2) \u00d7 AC = (1/2) \u00d7 3 = 3/2
\u800cDP = BC = \u221a3
\u2235 \u221a3 \u2260 3/2 \uff0c\u5373 DP \u4e0e DE \u4e0d\u91cd\u5408\u3001\u70b9P\u4e0e\u70b9E\u4e0d\u91cd\u5408\uff0c
\u2234 \u5f53\u70b9P\u5728\u8fd0\u52a8\u8fc7\u7a0b\u4e2d\u51fa\u73b0DP=BC\u65f6\uff0c \u6709\u4e24\u4e2a\u65f6\u523b\uff1a
\u2460 \u70b9P\u5c1a\u672a\u8d8a\u8fc7 \u70b9E \u524d\uff1b\u2461 \u70b9P\u8d8a\u8fc7 \u70b9E \u4e4b\u540e\u3002
\u2460 \u70b9P\u5c1a\u672a\u8d8a\u8fc7 \u70b9E \u524d\uff1a
\u5728 Rt\u25b3DPE \u4e2d\uff0c
sin\u2220DPE = DE / DP
= (3/2) / \u221a3
= \u221a3 / 2
\u800c sin60\u00b0 = \u221a3 / 2
\u2234 \u2220DPE = 60\u00b0
\u2234\u7531 \u201c\u4e09\u89d2\u5f62\u7684\u4e00\u4e2a\u5916\u89d2\u7b49\u4e8e\u4e0e\u5b83\u4e0d\u76f8\u90bb\u7684\u4e24\u4e2a\u5185\u89d2\u7684\u548c\u201d \u77e5\uff1a
\u2220DPE = \u2220DAP + \u2220PDA
\u2234\u2220PDA = \u2220DPE -- \u2220DAP
= 60\u00b0 -- 45\u00b0
= 15\u00b0
\u2461 \u70b9P\u8d8a\u8fc7 \u70b9E \u4e4b\u540e\uff1a
\u5728 Rt\u25b3DPE \u4e2d\uff0c
sin\u2220DPE = DE / DP
= (3/2) / \u221a3
= \u221a3 / 2
\u800c sin60\u00b0 = \u221a3 / 2
\u2234 \u2220DPE = 60\u00b0 \uff0c\u5373\uff1a\u2220DPA = 60\u00b0
\u5728 \u25b3DPA \u4e2d\uff0c\u7531\u4e09\u89d2\u5f62\u5185\u89d2\u548c\u5b9a\u7406\u5f97\uff1a
\u2220PDA = 180\u00b0 -- \u2220DPE -- \u2220DAP
= 180\u00b0 -- 60\u00b0 -- 45\u00b0
= 75\u00b0
\uff083\uff09\u9876\u70b9 \u201cQ\u201d \u6070\u597d\u5728\u8fb9BC\u4e0a\u3002\u60a8\u9898\u4e2d\u5c11\u6253\u4e86 Q \u3002
\u5f53\u70b9P\u8fd0\u52a8\u5230AC\u7684\u4e2d\u70b9\u5904\u65f6\uff0c
\u4ee5D\u3001P\u3001B\u3001Q\u4e3a\u9876\u70b9\u7684\u5e73\u884c\u56db\u8fb9\u5f62\u7684\u9876\u70b9Q\u6070\u597d\u5728\u8fb9BC\u4e0a\u3002\u7406\u7531\u5982\u4e0b\uff1a
\u2235 \u56db\u8fb9\u5f62DPBQ \u662f\u5e73\u884c\u56db\u8fb9\u5f62
\u2234 DP \u2016 BQ
\u800c BQ \u22a5 AC
\u2234 DP \u22a5 AC \u3002\u5373\uff1aDP\u662f\u7b49\u8170Rt\u25b3DAC\u7684\u5e95\u8fb9AC \u4e0a\u7684\u9ad8\u3002
\u2234 \u70b9P \u6b64\u65f6\u4e3a\u7ebf\u6bb5AC\u7684\u4e2d\u70b9\u3002\uff08\u7b49\u8170\u4e09\u89d2\u5f62\u5e95\u8fb9\u4e0a\u7684\u9ad8\u5e73\u5206\u5e95\u8fb9\uff09
\u2234\u5f53\u70b9P\u8fd0\u52a8\u5230AC\u7684\u4e2d\u70b9\u5904\u65f6\uff0c\u4ee5D\u3001P\u3001B\u3001Q\u4e3a\u9876\u70b9\u7684\u5e73\u884c\u56db\u8fb9\u5f62\u7684\u9876\u70b9Q\u6070\u597d\u5728\u8fb9BC\u4e0a\u3002
\u6c42\u6b64\u65f6\u5e73\u884c\u56db\u8fb9\u5f62DPBQ\u7684\u9762\u79ef\uff1a
\u4ee5 DP \u4e3a\u5e95\uff0c\u4ee5 DP \u4e0e BQ \u95f4\u7684 \u5782\u7ebf\u6bb5\u957f \u4e3a\u9ad8\u3002
DP \u4e0e BQ \uff08\u4e5f\u53ef\u4ee5\u8bf4DP \u4e0e BC\uff09\u95f4\u7684\u5782\u7ebf\u6bb5\u957f\u5373\u4e3aPC\u3002
\u2235 DP \u22a5 AC
\u2234 \u70b9P\u4e3aAC\u7684\u4e2d\u70b9
\u2234 PC = DP = AC/2 = 3/2
\u2234 S\u5e73\u884c\u56db\u8fb9\u5f62DPBQ = DP \u00d7 PC
= (3/2) \u00d7 (3/2)
= 9/4
（2）顶点坐标：（1/2，7/4）对称轴：直线X=1/2
（3）最小值：7/4
（4）抛物线和x轴不相交，和y轴交与（0，2）
（5）函数的图像：略
（6）当x为任何值都y>0
（7）x>1/2 时，y随x的增大而增大 x<1/2 时，y随x的增大而减小
1.向上
2.（-1/2，3/4） -1/2
3.最大：无限大 最小：3/4
4.与x轴不相交 y轴：3/4
5.自己画就好
6.y永远大於0
7.x小於1/2，y随x增大而减少，x小於1/2，y随x增大而增大
开口 向上
对称轴 -2A分之B 带入求定点坐标 配方 求最值
分别令 x和y 等于零 解方程 用5点法 画出图像
这些都说是 最基本的东西了 书上都有
绛旓細璋㈣阿銆 绁濆涔犺繘姝ワ紒鍒濅笁鏁板 鍥犲紡鍒嗚В x3-7x+6d鐨勨渪3鈥濇寚鐨勬槸涓夋鏂瑰惂锛 x^3-7x+6 =x^3-2x^2+x+2x^2-8x+6 =x(x-1)^2+(2x-6)(x-1) =(x-1)[x(x-1)+2x-6] =(x-1)(x+3)(x-2)鏈変竴閬鍒濅笁鏁板棰涓嶄細涓鍏冧簩娆 鍒嗚В鍥犲紡娉 (x+3a)(x-a)=0 ...
绛旓細24.瑙:(1)鏍规嵁棰鎰,c=3 鎵浠 瑙ｅ緱 鎵浠 鎶涚墿绾胯В鏋愬紡涓 (2)渚濋鎰忓彲寰桹A鐨勪笁绛夊垎鐐瑰垎鍒负(0,1),(0,2) 璁剧洿绾緾D鐨勮В鏋愬紡涓 褰撶偣D鐨勫潗鏍囦负(0,1)鏃,鐩寸嚎CD鐨勮В鏋愬紡涓 褰撶偣D鐨勫潗鏍囦负(0,2)鏃,鐩寸嚎CD鐨勮В鏋愬紡涓 (3)濡傚浘,鐢遍鎰,鍙緱 鐐筂鍏充簬x杞寸殑瀵圭О鐐逛负 鐐笰鍏充簬鎶涚墿绾垮绉拌酱 鐨...
绛旓細133^5 + 110^5 + 84^5 + 27^5 = n^5 61917364224 鏄剧劧n涓3浣嶆暟锛屼笖棣栦綅涓1 3^5 灏炬暟 3 0^5 灏炬暟 0 4^5 灏炬暟 4 7^5 灏炬暟 7 鎵浠鐨勫熬鏁皔^5 鐨勫熬鏁颁负 3 + 0 + 7 + 4 = 14 鐨勫熬鏁 4锛寉 = 4, n鍙互琛ㄧず涓 1x4 涓変綅鏁 鍥犱负4涓暟涓渶澶х殑涓 133锛 鎵浠 > 3锛...
绛旓細杩欐槸涓寑鍑忛熺洿绾胯繍鍔ㄩ棶棰樸45绫虫槸杩愬姩鐨勮矾绋嬶紱10绉掓槸杩愬姩鐨勬椂闂达紱1锛2涓ら棶鏄眰杩愬姩鐨勫垵閫熷害鍜屽姞閫熷害锛堝钩鍧囨瘡绉掑噺灏戝灏戯級锛屽湪姝ゅ熀纭涓婂彲浠ユ眰鍑虹涓夐棶銆傝В娉曚竴锛氬寑閫熺洿绾胯繍鍔ㄧ殑鍏紡锛歴=v0*t+(1/2)a*t^2锛泇=v0+a*t 鍏朵腑v0鏄垵閫熷害锛0鏄笅鏍囷級v鏄湯閫熷害锛沘鏄姞閫熷害锛泃鏄椂闂淬傛湰棰樻槸...
绛旓細1.锛1锛2x-1=鏍瑰彿3 鎴 -鏍瑰彿3 x=锛堟牴鍙3+1锛/2 鎴 锛-鏍瑰彿3+1锛/2 锛2锛墄^2=25 x=5鎴-5 (3)(3x-6)(x-2)=x(x-2)(2x-6)(x-2)=0 2x-6=0鎴 x-2=0 x=3鎴2 锛4锛 2x^2-11x+2=0 delta=(-11)^2-4*2*2=105 x=锛11+鏍瑰彿105锛/4 鎴 锛11-鏍瑰彿105锛/4...
绛旓細鍒嗘瀽锛氾紙1锛夌敱棰樻剰鎶涚墿绾縴=ax²+bx+c锛坅锛0锛夌殑鍥捐薄缁忚繃鐐笲锛12锛0锛夊拰C锛0锛-6锛夛紝瀵圭О杞翠负x=2锛屾牴鎹緟瀹氱郴鏁版硶鍙互姹傚緱璇ユ姏鐗╃嚎鐨勮В鏋愬紡锛涘亣璁惧瓨鍦紝璁惧嚭鏃堕棿t锛屽垯鏍规嵁绾挎PQ琚洿绾緾D鍨傜洿骞冲垎锛屽啀鐢卞瀭鐩村钩鍒嗙嚎鐨勬ц川鍙婂嬀鑲″畾鐞嗘潵姹傝Вt锛岀湅t鏄惁瀛樺湪锛涳紙2锛夊亣璁剧洿绾縳=1涓婃槸瀛樺湪鐐筂...
绛旓細璇佹槑锛氬洜涓築P骞冲垎鈭燗BC,鎵浠モ垹PBC=鈭燗BC/2,鍚岀悊鈭燩CB=鈭燗CB/2 鎵浠モ垹PBC+鈭燩CB=鈭燗BC/2+鈭燗CB/2=(1/2)(鈭燗BC+鈭燗CB)鍦ㄤ笁瑙掑舰BCP涓紝鈭燘PC=180-鈭燩BC-鈭燩CB=180-(鈭燩BC+鈭燩CB)=180-(1/2)(鈭燗BC+鈭燗CB)鍙堝湪鈻矨BC涓紝鈭燗+鈭燗BC+鈭燗CB=180,鈭燗BC+鈭燗CB=鈭180-鈭燗 鎵浠モ垹BPC=...
绛旓細瑙ｇ瓟锛氾紙1锛夊綋P鐐硅繍鍔ㄥ埌AB鐨勪腑鐐规椂锛岃嫢鎭板ソPQ//BC鏃讹紝鏍规嵁鍕捐偂瀹氱悊锛屽彲寰楋細AC=鈭氾紙AB²+BC²锛=5锛汚P=AB/2=4锛泃=AP/1=4(m/s)锛涘張鍙緱锛歺=AC/t=5/4=1.25(m/s锛夛紙2锛1銆嬩笌绗竴棰樿В绛旂浉鍚屻傦紙鍥句竴锛2銆嬫牴鎹嬀鑲″畾鐞嗭紝鍙緱 锛堝浘浜岋級 AP鐨勯熷害涓嶢Q鐨勯熷害鐨勬瘮鏄5...
绛旓細棰樺共 宸茬煡鎶涚墿绾縴=ax2+bx缁忚繃鐐笰锛-3锛-3锛夊拰鐐筆锛坱锛0锛夛紝涓攖鈮0锛庯紙1锛夎嫢璇ユ姏鐗╃嚎鐨勫绉拌酱缁忚繃鐐笰锛屽鍥撅紝璇烽氳繃瑙傚療鍥捐薄锛屾寚鍑烘鏃秠鐨勬渶灏忓硷紝骞跺啓鍑簍鐨勫硷紱锛2锛夎嫢t=-4锛屾眰a銆乥鐨勫硷紝骞舵寚鍑烘鏃舵姏鐗╃嚎鐨勫紑鍙ｆ柟鍚戯紱锛3锛夌洿鎺ュ啓鍑轰娇璇ユ姏鐗╃嚎寮鍙ｅ悜涓嬬殑t鐨勪竴涓硷紟绛旀 瑙ｏ細锛1锛夆埖...
绛旓細鈶 w=-2x²+120x-1600 =-2 (x-30) ²+200,鈭村綋x=30鏃,y鏈夋渶澶у200. 鈭村綋閿鍞环瀹氫负30鍏/鍗冨厠鏃,姣忓ぉ鍙幏鏈澶ч攢鍞埄娑200鍏. 鈶 褰搚=150鏃,鍙緱鏂圭▼ -2 (x-30 )2 +200=150.瑙ｈ繖涓柟绋,寰 x1=25,x2=3 5. 鏍规嵁棰鎰,x2=35涓嶅悎棰樻剰,搴旇垗鍘...
