求解一个有关实数的数学题。 一道有关实数的数学题找规律
\u5173\u4e8e\u521d\u4e00\u6570\u5b66\u5b9e\u6570\u7684\u5e94\u7528\u9898\u53ca\u7b54\u6848\uff085\u9053\uff09\uff08\u6025\u901f\uff09\u62dc\u6258\u4e861.\u5df2\u77e5\u56db\u8fb9\u5f62ABCD\u4e3a\u4efb\u610f\u51f8\u56db\u8fb9\u5f62\uff0cE\u3001F\u3001G\u3001H\u5206\u522b\u662f\u8fb9AB\u3001BC\u3001CD\u3001DA\u7684\u4e2d\u70b9\uff0c\u7528S\u3001P\u5206\u522b\u8868\u793a\u56db\u8fb9\u5f62ABCD\u7684\u9762\u79ef\u548c\u5468\u957f\uff1bS1\u3001P1\u5206\u522b\u8868\u793a\u56db\u8fb9\u5f62EFGH\u7684\u9762\u79ef\u548c\u5468\u957f\uff0e\u8bbeK
= SS1\uff0cK1 = PP1 \uff0c\u5219\u4e0b\u9762\u5173\u4e8eK\u3001K1\u7684\u8bf4\u6cd5\u6b63\u786e\u7684\u662f \uff08 \uff09\uff0e
A.K\u3001K1\u5747\u4e3a\u5e38\u503c B.K\u4e3a\u5e38\u503c\uff0cK1\u4e0d\u4e3a\u5e38\u503c
C.K\u4e0d\u4e3a\u5e38\u503c,K1\u4e3a\u5e38\u503c D.K\u3001K1\u5747\u4e0d\u4e3a\u5e38\u503c
2.\u5df2\u77e5m\u4e3a\u5b9e\u6570\uff0c\u4e14sin\u03b1\u3001cos\u03b1\u662f\u5173\u4e8ex\u7684\u65b9\u7a0b3x2 \u2013mx + 1 =
0\u7684\u4e24\u6839\uff0c\u5219sin4\u03b1+ cos4\u03b1\u7684\u503c\u4e3a \uff08 \uff09\uff0e
A.29 B. 13 C. 79 D.1
3.\u5173\u4e8ex\u7684\u65b9\u7a0b|x2x\u20131 |=
a\u4ec5\u6709\u4e24\u4e2a\u4e0d\u540c\u7684\u5b9e\u6839\uff0c\u5219\u5b9e\u6570a\u7684\u53d6\u503c\u8303\u56f4\u662f \uff08 \uff09\uff0e
A.a > 0 B.a \u22654 C.2 < a < 4 D.0 < a
< 4
4.\u8bbeb>0\uff0ca2 -2ab + c2 = 0\uff0cbc > a2\uff0c\u5219\u5b9e\u6570a\u3001b\u3001c\u7684\u5927\u5c0f\u5173\u7cfb\u662f \uff08 \uff09\uff0e
A.b
> c >a B.c> a > b C.a > b > c D.b > a > c
5.\u8bbea\u3001b\u4e3a\u6709\u7406\u6570\uff0c\u4e14\u6ee1\u8db3\u7b49\u5f0fa + b3 =6 ⋅1 + 4 + 23 \uff0c\u5219a + b\u7684\u503c\u4e3a
\uff08 \uff09\uff0e
A.2 B.4 C.6
D.8
6.\u5c06\u6ee1\u8db3\u6761\u4ef6\u201c\u81f3\u5c11\u51fa\u73b0\u4e00\u4e2a\u6570\u5b570\uff0c\u4e14\u662f4\u7684\u500d\u6570\u7684\u6b63\u6574\u6570\u201d\u4ece\u5c0f\u5230\u5927\u6392\u6210\u4e00\u5217\u6570\uff1a20\uff0c40\uff0c60\uff0c80\uff0c100\uff0c104\uff0c\u2026\u2026\uff0c\u5219\u8fd9\u5217\u6570\u4e2d\u7684\u7b2c158\u4e2a\u6570\u4e3a
\uff08 \uff09\uff0e
A.2000 B.2004 C.2008 D.2012
\u4e8c\u3001\u586b\u7a7a\u9898\uff08\u672c\u9898\u6ee1\u520628\u5206,\u6bcf\u5c0f\u98987\u5206\uff09
1.\u51fd\u6570y = x2
-2006|x|+ 2008\u7684\u56fe\u8c61\u4e0ex\u8f74\u4ea4\u70b9\u7684\u6a2a\u5750\u6807\u4e4b\u548c\u7b49\u4e8e__________\uff0e
2.\u5728\u7b49\u8170Rt\u25b3ABC\u4e2d\uff0cAC = BC =1\uff0c M\u662fBC\u7684\u4e2d\u70b9\uff0c
CE\u22a5AM\u4e8eE\u4ea4AB\u4e8eF\uff0c\u5219S\u22bfMBF = __________.
3.\u4f7fx2 + 4 + (8 - x)2 + 16
\u53d6\u6700\u5c0f\u503c\u7684\u5b9e\u6570x\u7684\u503c\u4e3a__________.
4.\u5728\u5e73\u9762\u76f4\u89d2\u5750\u6807\u7cfb\u4e2d\uff0c\u6b63\u65b9\u5f62OABC\u7684\u9876\u70b9\u5750\u6807\u5206\u522b\u4e3aO(0,0)\uff0cA(100,0)\uff0cB(100,100)\uff0cC(0,100)\uff0e\u82e5\u6b63\u65b9\u5f62OABC\u5185\u90e8\uff08\u8fb9\u754c\u53ca\u9876\u70b9\u9664\u5916\uff09\u4e00\u683c\u70b9P\u6ee1\u8db3\uff1a
S\u22bfPOA ⋅ S\u22bfPBC = S\u22bfPAB⋅S\u22bfPOC \uff0c\u5c31\u79f0\u683c\u70b9P\u4e3a\u201c\u597d\u70b9\u201d\uff0c\u5219\u6b63\u65b9\u5f62OABC\u5185\u90e8\u201c\u597d\u70b9\u201d\u7684\u4e2a\u6570\u4e3a__________.
\uff08\u6ce8\uff1a\u6240\u8c13\u201c\u683c\u70b9\u201d\uff0c\u662f\u6307\u5728\u5e73\u9762\u76f4\u89d2\u5750\u6807\u7cfb\u4e2d\u6a2a\u3001\u7eb5\u5750\u6807\u5747\u4e3a\u6574\u6570\u7684\u70b9\uff0e\uff09
2006\u5e74\u5168\u56fd\u521d\u4e2d\u6570\u5b66\u8054\u5408\u7ade\u8d5b\u8bd5\u5377(\u8fbd\u5b81\uff09
\u7b2c\u4e8c\u8bd5
(4\u67089\u65e5 \u4e0a\u5348 10:00-11:30)
\u4e00\u3001(\u672c\u9898\u6ee1\u520620\u5206)
\u5df2\u77e5\u5173\u4e8ex\u7684\u4e00\u5143\u4e8c\u6b21\u65b9\u7a0bx2 +2(a + 2b + 3)x+(a2
+ 4b2 + 99)= 0\u65e0\u76f8\u5f02\u4e24\u5b9e\u6839\uff0c\u5219\u6ee1\u8db3\u6761\u4ef6\u7684\u6709\u5e8f\u6b63\u6574\u6570\u7ec4(a,b)\u6709\u591a\u5c11\u7ec4\uff1f
\u4e8c\u3001(\u672c\u9898\u6ee1\u520625\u5206)
\u5982\u56fe\uff0cD\u4e3a\u7b49\u8170\u25b3ABC\u5e95\u8fb9BC\u7684\u4e2d\u70b9\uff0cE\u3001F\u5206\u522b\u4e3aAC\u53ca\u5176\u5ef6\u957f\u7ebf
\u4e0a\u7684\u70b9\uff0e\u53c8\u5df2\u77e5\u2220EDF = 90\u00b0\uff0cED = DF = 1\uff0cAD =
5\uff0e\u6c42\u7ebf\u6bb5BC\u7684\u957f\uff0e
\u4e09\u3001(\u672c\u9898\u6ee1\u520625\u5206)
\u5728\u5e73\u884c\u56db\u8fb9\u5f62ABCD\u4e2d\uff0c\u2220A\u7684\u5e73\u5206\u7ebf\u5206\u522b\u4e0eBC\u53caDC\u7684\u5ef6\u957f
\u7ebf\u4ea4\u4e8e\u70b9E\u3001F\uff0c\u70b9O\u3001O1\u5206\u522b\u4e3a\u25b3CEF\u3001\u25b3ABE\u7684\u5916\u5fc3\uff0e (1)\u6c42\u8bc1\uff1a
O\u3001E\u3001O1\u4e09\u70b9\u5171\u7ebf\uff1b (2)\u6c42\u8bc1\uff1a\u82e5\u2220ABC =
70\u00b0\uff0c\u6c42\u2220OBD\u7684
\u5ea6\u6570\uff0e
\u53c2\u8003\u7b54\u6848\uff1a
\u9009\u62e9\u9898\uff1aBCDABC
\u586b\u7a7a\u9898\uff1a1. 0 2. 112 3. 38 4.
197
\u89e3\u7b54\u9898\uff1a1. 16 2. 107 3. (1)
\u8bc1\u660e\u76f8\u4f3c\u4e09\u89d2\u5f62\u7684\u5bf9\u5e94\u89d2\u76f8\u7b49\uff1b\uff082\uff0935\u00b0.
\u24601/3x=-4
x=-12
\u24616x-a=0
x=a/6
\u2460\u7684\u89e3\u6bd4\u2461\u7684\u89e3\u59275
\u6240\u4ee5-12-a/6=5
a/6=-17
a=-102
\u2462x/a-2/51=0
x/(-102)-2/51=0
x/102=-2/51
x=-4
\u89e3\uff1a\u221a[4-(4/17)]=4\u221a(4/17)
\u89c4\u5f8b\u662f\uff1a
\u221a{n-[(n/(n²+1)]}=n\u221a[(n/(n²+1)]
设x是有理数，用[x]表示不超过x的最大整数
观察代数式：x=a1+a2/2!+a3/3!+...+ak/k!
很容易看出
a1=[x]
a2=[x2]，其中：x2=(x-a1)*2
a3=[x3]，其中：x3=(x2-a2)*3
a4=[x4]，其中：x4=(x3-a3)*4
a5=[x5]，其中：x5=(x4-a4)*5
...
由于x是有理数
所以x2，x3，...都是有理数
又每次迭代xi都是有理数(x(i-1)-a(i-1))乘以整数i
所以最终必定有一个xk成为整数
即：ak=[xk]=xk
于是：xk-ak=0
后面的迭代就全是0了
得到：x=a1+a2/2!+a3/3!+...+ak/k!
若说几率的话那么便是A了因为它既是众数又是中数并占了五分之三的几率但ABCDE分别有五分之一被抽到的几率所以这道题并没有答案看运气了 ABCDE都是有可能的
1.本题考恒等概念。
(1)a=b=c=0.
(2)a:b=α:β，c=0.
(3)α：β：γ=A:B:C.
绛旓細绗4棰 鍘熷紡=鈭(-2)²-鈭(2鍙1/4)-³鈭27 =鈭4-鈭(9/4)-3 =2-(3/2)-3 =-5/2
绛旓細1.鈭祒銆亂鏄鏁癮鐨勪袱涓钩鏂规牴 鈭磝+y=0 鍙堚埖2x-y=2 鈭磞=2x-2 鈭磝+2x-2=0 鈭3x=2 鈭磝=3锛2 鈭碼=锛3/2锛²=9锛4 2.鈭(鏍瑰彿5)鐨勮В涓猴細2.236067...鎵浠ュ皬鏁伴儴鍒嗕负锛(鏍瑰彿5)-2=a 鍙堚埖5-(鏍瑰彿5)鐨勮В涓猴細2.7639320...鎵浠ュ皬鏁伴儴鍒嗕负5-锛堟牴鍙5锛-2 =3-锛堟牴鍙5...
绛旓細绛旓細|2008-a|+鈭(a-2009)=a>0 鍥犱负锛歛-2009>=0 鎵浠ワ細a>=2009 鎵浠ュ師寮忓寲涓猴細a-2008+鈭(a-2009)=a 鎵浠ワ細鈭(a-2009)=2008 鎵浠ワ細a-2009=2008²鎵浠ワ細a-2008²=2009
绛旓細鎵浠锛4锛孊=9 鎵浠モ垰AB=鈭4*9=6 2x^2-4x+2+鈭氾紙2x+2y-1锛夛紳0 璁2X^2-4X+2=0涓擷+2Y-1=0锛屽垯2X²-4X+2+鏍瑰彿涓嬶紙X+2Y-1锛=0銆傚緱鍒癤=1锛孻=0锛屾墍浠Y=0
绛旓細101*102=(100+1)102=10200+102=10302 100*103=10300 100*101*102*103+1=10300*10302+1=(10301-1)(10301+1)+1=10301^2-1+1=10301^2 鎵浠 绛旀鏄10301
绛旓細1銆佸洜涓轰笁娆℃牴鍙3y-1鍜屼笁娆℃牴鍙1-x浜掍负鐩稿弽鏁 鎵浠ヤ袱杈逛笁娆℃柟寰3y-1=-锛1-x锛夎В寰梮鍒嗕箣y鐨勫=1/3 2銆佸洜涓瀹炴暟x y 婊¤冻 鏍瑰彿2x-3y-1 + 缁濆鍊紉-2y+2=0 鎵浠2x-3y-1=0 x-2y+2=0 瑙ｆ浜屽厓涓娆℃柟绋嬪緱x=8 y=5 灏唜=5浠ｅ叆 2x-4鍒嗕箣5鐨勫钩鏂规牴=6鍒嗕箣1鍊嶆牴鍙15銆
绛旓細(x+1)(x-1)+2(x+1)=x^2-1+2x+2 =x^2+2x+1 褰搙=鏍瑰彿2鏃讹紝鍘熷紡=x^2+2x+1 =(鏍瑰彿2)^2+2(鏍瑰彿2)+1 =2+2鍊嶆牴鍙2+1 =3+2鍊嶆牴鍙2
绛旓細|鏍瑰彿2+鏍瑰彿5|=鏍瑰彿2+鏍瑰彿5锛屾墍浠ュ師寮=锛堟牴鍙5-鏍瑰彿2锛-锛堟牴鍙5+鏍瑰彿2锛=鏍瑰彿5-鏍瑰彿2-鏍瑰彿2-鏍瑰彿5 =-2鏍瑰彿2 绾︾瓑浜-2.83 锛2锛夊洜涓烘牴鍙2锛渁锛滄淳 鎵浠-娲撅紲0锛屾牴鍙2-娲撅紲0锛屾墍浠a-娲緗=娲-a锛寍鏍瑰彿2-a|=a-鏍瑰彿2 鎵浠ュ師寮=娲-a+a-鏍瑰彿2=娲-鏍瑰彿2绾︾瓑浜1.73 ...
绛旓細锛堟牴鍙凤級a-3 + 锛堢粷瀵瑰硷級b-4 + c锛堝钩鏂癸級-10c + 25 = 0锛岋紙鏍瑰彿锛塧-3 + 锛堢粷瀵瑰硷級b-4 + (c-5)锛堝钩= 0锛宎-3=0,b-4=0,c-5=0 a=3,b=4,c=5 c^2=a^2+b^2 鈻矨BC鏄洿瑙掍笁瑙掑舰
绛旓細1-2鈭 3 =1-鈭 2鈭 6 > 1-3鈭 2=1-鈭 3鈭 6 鑻涓涓姝ｆ暟鐨勫钩鏂规牴鏄痑+1鍜宎-3,鍒欒繖涓鏁版槸澶氬皯(杩欎釜棰樼洰鍏ㄥ悧)璇村嚭鐢ㄥ嚑浣曚綔鍥炬硶琛ㄧず鏁拌酱涓婄殑鈭 5鍜-鈭 10鐨勬柟娉曪細鐢诲嚭闀夸负2锛屽涓1鐨勯暱鏂瑰舰锛堢煩褰級锛屽瑙掔嚎闀垮害灏辨槸鈭 5锛岀敤鍦嗚鍦ㄦ暟杞翠笂閲忓嚭鍚屾牱鐨勯暱搴︼紱鍦ㄦ暟杞村弽鏂瑰悜锛岀敾鍑...
