高一数学题 高一数学题（函数）？
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1\uff0e\u4e0b\u5217\u516b\u4e2a\u5173\u7cfb\u5f0f\u2460{0}= \u2461 =0 \u2462 { } \u2463 { } \u2464{0} \u24650 \u2466 {0} \u2467 { }\u5176\u4e2d\u6b63\u786e\u7684\u4e2a\u6570\uff08 \uff09
\uff08A\uff094 \uff08B\uff095 \uff08C\uff096 \uff08D\uff097
2\uff0e\u96c6\u5408{1\uff0c2\uff0c3}\u7684\u771f\u5b50\u96c6\u5171\u6709\uff08 \uff09
\uff08A\uff095\u4e2a \uff08B\uff096\u4e2a \uff08C\uff097\u4e2a \uff08D\uff098\u4e2a
3\uff0e\u96c6\u5408A={x } B={ } C={ }\u53c8 \u5219\u6709\uff08 \uff09
\uff08A\uff09\uff08a+b\uff09 A (B) (a+b) B (C)(a+b) C (D) (a+b) A\u3001B\u3001C\u4efb\u4e00\u4e2a
4\uff0e\u8bbeA\u3001B\u662f\u5168\u96c6U\u7684\u4e24\u4e2a\u5b50\u96c6\uff0c\u4e14A B\uff0c\u5219\u4e0b\u5217\u5f0f\u5b50\u6210\u7acb\u7684\u662f\uff08 \uff09
\uff08A\uff09CUA CUB \uff08B\uff09CUA CUB=U
\uff08C\uff09A CUB= \uff08D\uff09CUA B=
5\uff0e\u5df2\u77e5\u96c6\u5408A={ } B={ }\u5219A =\uff08 \uff09
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\uff08C\uff09{ } \uff08D\uff09{ }
6\uff0e\u4e0b\u5217\u8bed\u53e5\uff1a\uff081\uff090\u4e0e{0}\u8868\u793a\u540c\u4e00\u4e2a\u96c6\u5408\uff1b\uff082\uff09\u75311\uff0c2\uff0c3\u7ec4\u6210\u7684\u96c6\u5408\u53ef\u8868\u793a\u4e3a{1\uff0c2\uff0c3}\u6216{3\uff0c2\uff0c1}\uff1b\uff083\uff09\u65b9\u7a0b\uff08x-1\uff092(x-2)2=0\u7684\u6240\u6709\u89e3\u7684\u96c6\u5408\u53ef\u8868\u793a\u4e3a{1\uff0c1\uff0c2}\uff1b\uff084\uff09\u96c6\u5408{ }\u662f\u6709\u9650\u96c6\uff0c\u6b63\u786e\u7684\u662f\uff08 \uff09
\uff08A\uff09\u53ea\u6709\uff081\uff09\u548c\uff084\uff09 \uff08B\uff09\u53ea\u6709\uff082\uff09\u548c\uff083\uff09
\uff08C\uff09\u53ea\u6709\uff082\uff09 \uff08D\uff09\u4ee5\u4e0a\u8bed\u53e5\u90fd\u4e0d\u5bf9
7\uff0e\u5df2\u77e5A={1\uff0c2\uff0ca2-3a-1},B={1,3},A {3,1}\u5219a\u7b49\u4e8e\uff08 \uff09
\uff08A\uff09-4\u62161 \uff08B\uff09-1\u62164 \uff08C\uff09-1 \uff08D\uff094
8.\u8bbeU={0\uff0c1\uff0c2\uff0c3\uff0c4}\uff0cA={0\uff0c1\uff0c2\uff0c3}\uff0cB={2\uff0c3\uff0c4}\uff0c\u5219\uff08CUA\uff09 \uff08CUB\uff09=\uff08 \uff09
\uff08A\uff09{0} \uff08B\uff09{0\uff0c1}
\uff08C\uff09{0\uff0c1\uff0c4} \uff08D\uff09{0\uff0c1\uff0c2\uff0c3\uff0c4}
9\uff0e\u8bbeS\u3001T\u662f\u4e24\u4e2a\u975e\u7a7a\u96c6\u5408\uff0c\u4e14S T\uff0cT S\uff0c\u4ee4X=S \u90a3\u4e48S X=\uff08 \uff09
\uff08A\uff09X \uff08B\uff09T \uff08C\uff09 \uff08D\uff09S
10\uff0e\u8bbeA={x },B={x },\u82e5A B={2,3,5},A\u3001B\u5206\u522b\u4e3a\uff08 \uff09
\uff08A\uff09{3\uff0c5}\u3001{2\uff0c3} \uff08B\uff09{2\uff0c3}\u3001{3\uff0c5}
\uff08C\uff09{2\uff0c5}\u3001{3\uff0c5} \uff08D\uff09{3\uff0c5}\u3001{2\uff0c5}
11\uff0e\u8bbe\u4e00\u5143\u4e8c\u6b21\u65b9\u7a0bax2+bx+c=0(a<0)\u7684\u6839\u7684\u5224\u522b\u5f0f \uff0c\u5219\u4e0d\u7b49\u5f0fax2+bx+c 0\u7684\u89e3\u96c6\u4e3a\uff08 \uff09
\uff08A\uff09R \uff08B\uff09
\uff08C\uff09{ } \uff08D\uff09{ }
\uff08A\uff09P Q
\uff08B\uff09Q P
\uff08C\uff09P=Q \uff08D\uff09P Q=
12\uff0e\u5df2\u77e5P={ }\uff0cQ={ \uff0c\u5bf9\u4e8e\u4e00\u5207 R\u6210\u7acb}\uff0c\u5219\u4e0b\u5217\u5173\u7cfb\u5f0f\u4e2d\u6210\u7acb\u7684\u662f\uff08 \uff09
13\uff0e\u82e5M={ }\uff0cN={ Z}\uff0c\u5219M N\u7b49\u4e8e\uff08 \uff09
\uff08A\uff09 \uff08B\uff09{ } \uff08C\uff09{0} \uff08D\uff09Z
14\uff0e\u4e0b\u5217\u5404\u5f0f\u4e2d\uff0c\u6b63\u786e\u7684\u662f\uff08 \uff09
\uff08A\uff092
\uff08B\uff09{ }
\uff08C\uff09{ }
\uff08D\uff09{ }={ }
15\uff0e\u8bbeU={1\uff0c2\uff0c3\uff0c4\uff0c5}\uff0cA\uff0cB\u4e3aU\u7684\u5b50\u96c6\uff0c\u82e5A B={2}\uff0c\uff08CUA\uff09 B={4}\uff0c\uff08CUA\uff09 \uff08CUB\uff09={1\uff0c5}\uff0c\u5219\u4e0b\u5217\u7ed3\u8bba\u6b63\u786e\u7684\u662f\uff08 \uff09
\uff08A\uff093 \uff08B\uff093
\uff08C\uff093 \uff08D\uff093
16\uff0e\u82e5U\u3001 \u5206\u522b\u8868\u793a\u5168\u96c6\u548c\u7a7a\u96c6\uff0c\u4e14\uff08CUA\uff09 A\uff0c\u5219\u96c6\u5408A\u4e0eB\u5fc5\u987b\u6ee1\u8db3\uff08 \uff09
(A) (B)
(C)B= (D)A=U\u4e14A B
17\uff0e\u5df2\u77e5U=N\uff0cA={ }\uff0c\u5219CUA\u7b49\u4e8e\uff08 \uff09
\uff08A\uff09{0\uff0c1\uff0c2\uff0c3\uff0c4\uff0c5\uff0c6} \uff08B\uff09{1\uff0c2\uff0c3\uff0c4\uff0c5\uff0c6}
\uff08C\uff09{0\uff0c1\uff0c2\uff0c3\uff0c4\uff0c5} \uff08D\uff09{1\uff0c2\uff0c3\uff0c4\uff0c5}
18\uff0e\u4e8c\u6b21\u51fd\u6570y=-3x2+mx+m+1\u7684\u56fe\u50cf\u4e0ex\u8f74\u6ca1\u6709\u4ea4\u70b9\uff0c\u5219m\u7684\u53d6\u503c\u8303\u56f4\u662f\uff08 \uff09
\uff08A\uff09{ } \uff08B\uff09{ }
\uff08C\uff09{ } \uff08D\uff09{ }
19\uff0e\u8bbe\u5168\u96c6U={\uff08x,y\uff09 },\u96c6\u5408M={\uff08x,y\uff09 }\uff0cN={(x,y) },\u90a3\u4e48\uff08CUM\uff09 \uff08CUN\uff09\u7b49\u4e8e\uff08 \uff09
\uff08A\uff09{\uff082\uff0c-2\uff09} \uff08B\uff09{\uff08-2\uff0c2\uff09}
\uff08C\uff09 \uff08D\uff09\uff08CUN\uff09
20\uff0e\u4e0d\u7b49\u5f0f <x2-4\u7684\u89e3\u96c6\u662f\uff08 \uff09
\uff08A\uff09{x } \uff08B\uff09{x }
\uff08C\uff09{ x } \uff08D\uff09{ x }
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1\uff0e \u5728\u76f4\u89d2\u5750\u6807\u7cfb\u4e2d\uff0c\u5750\u6807\u8f74\u4e0a\u7684\u70b9\u7684\u96c6\u5408\u53ef\u8868\u793a\u4e3a
2\uff0e \u82e5A={1,4,x},B={1,x2}\u4e14A B=B\uff0c\u5219x=
3\uff0e \u82e5A={x } B={x },\u5168\u96c6U=R\uff0c\u5219A =
4\uff0e \u82e5\u65b9\u7a0b8x2+(k+1)x+k-7=0\u6709\u4e24\u4e2a\u8d1f\u6839\uff0c\u5219k\u7684\u53d6\u503c\u8303\u56f4\u662f
5\uff0e \u96c6\u5408{a,b,c}\u7684\u6240\u6709\u5b50\u96c6\u662f \u771f\u5b50\u96c6\u662f \uff1b\u975e\u7a7a\u771f\u5b50\u96c6\u662f
6\uff0e \u65b9\u7a0bx2-5x+6=0\u7684\u89e3\u96c6\u53ef\u8868\u793a\u4e3a
\u65b9\u7a0b\u7ec4
7\uff0e\u8bbe\u96c6\u5408A={ },B={x },\u4e14A B\uff0c\u5219\u5b9e\u6570k\u7684\u53d6\u503c\u8303\u56f4\u662f
\u3002
8\uff0e\u8bbe\u5168\u96c6U={x \u4e3a\u5c0f\u4e8e20\u7684\u975e\u8d1f\u5947\u6570}\uff0c\u82e5A \uff08CUB\uff09={3\uff0c7\uff0c15}\uff0c\uff08CUA\uff09 B={13\uff0c17\uff0c19}\uff0c\u53c8\uff08CUA\uff09 \uff08CUB\uff09= \uff0c\u5219A B=
9\uff0e\u8bbeU={\u4e09\u89d2\u5f62}\uff0cM={\u76f4\u89d2\u4e09\u89d2\u5f62}\uff0cN={\u7b49\u8170\u4e09\u89d2\u5f62}\uff0c\u5219M N=
M N= CUM=
CUN= CU\uff08M N\uff09=
10\uff0e\u8bbe\u5168\u96c6\u4e3a \uff0c\u7528\u96c6\u5408A\u3001B\u3001C\u7684\u4ea4\u3001\u5e76\u3001\u8865\u96c6\u7b26\u53f7\u8868\u56fe\u4e2d\u7684\u9634\u5f71\u90e8\u5206\u3002
\uff081\uff09 \uff082\uff09
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\u4e09\u3001\u89e3\u7b54\u9898
1\uff0e\u8bbe\u5168\u96c6U={1\uff0c2\uff0c3\uff0c4}\uff0c\u4e14={ x2-5x+m=0,x U}\u82e5CUA={1\uff0c4}\uff0c\u6c42m\u7684\u503c\u3002
2\uff0e\u5df2\u77e5\u96c6\u5408A={a \u5173\u4e8ex\u7684\u65b9\u7a0bx2-ax+1=0,\u6709\u5b9e\u6839}\uff0cB={a \u4e0d\u7b49\u5f0fax2-x+1>0\u5bf9\u4e00\u5207x R\u6210\u7acb},\u6c42A B\u3002
3\uff0e\u5df2\u77e5\u96c6\u5408A={a2,a+1,-3},B={a-3,2a-1,a2+1}, \u82e5A B={-3}\uff0c\u6c42\u5b9e\u6570a\u3002
4\uff0e\u5df2\u77e5\u65b9\u7a0bx2-(k2-9)+k2-5k+6=0\u7684\u4e00\u6839\u5c0f\u4e8e1\uff0c\u53e6\u4e00\u6839\u5927\u4e8e2\uff0c\u6c42\u5b9e\u6570k\u7684\u53d6\u503c\u8303\u56f4\u3002
5\uff0e\u8bbeA={x ,\u5176\u4e2dx R,\u5982\u679cA B=B\uff0c\u6c42\u5b9e\u6570a\u7684\u53d6\u503c\u8303\u56f4\u3002
6\uff0e\u8bbe\u5168\u96c6U={x },\u96c6\u5408A={x },B={ x2+px+12=0},\u4e14\uff08CUA\uff09 B={1\uff0c4\uff0c3\uff0c5}\uff0c\u6c42\u5b9e\u6570P\u3001q\u7684\u503c\u3002
7\uff0e\u82e5\u4e0d\u7b49\u5f0fx2-ax+b0\u7684\u89e3\u96c6\u3002
8\uff0e\u96c6\u5408A={\uff08x,y\uff09 },\u96c6\u5408B={\uff08x,y\uff09 ,\u4e140 }\uff0c\u53c8A \uff0c\u6c42\u5b9e\u6570m\u7684\u53d6\u503c\u8303\u56f4\u3002
\u7b2c\u4e00\u5355\u5143 \u96c6\u5408
\u4e00\u3001 \u9009\u62e9\u9898
\u9898\u53f7 1 2 3 4 5 6 7 8 9 10
\u7b54\u6848 B C B C B C B C D A
\u9898\u53f7 11 12 13 14 15 16 17 18 19 20
\u7b54\u6848 D A A D C D A D A B
\u4e8c\u3001 \u586b\u7a7a\u9898\u7b54\u6848
1\uff0e{(x,y) } 2.0, 3.{x ,\u6216x 3} 4.{ } 5. ,{a},{b},{c},{a,b},{a,c},{b,c},{a,b,c};\u9664\u53bb{a,b,c}\u5916\u6240\u6709\u5b50\u96c6\uff1b\u9664\u53bb \u53ca{a,b,c}\u5916\u7684\u6240\u6709\u5b50\u96c6 6.{2,3};{2,3} 7.{ } 8.{1,5,9,11} 9.{\u7b49\u8170\u76f4\u89d2\u4e09\u89d2\u5f62}\uff1b{\u7b49\u8170\u6216\u76f4\u89d2\u4e09\u89d2\u5f62}\uff0c{\u659c\u4e09\u89d2\u5f62}\uff0c{\u4e0d\u7b49\u8fb9\u4e09\u89d2\u5f62}\uff0c{\u65e2\u975e\u7b49\u8170\u4e5f\u975e\u76f4\u89d2\u4e09\u89d2\u5f62}\u3002 10.\uff081\uff09 \uff08A B\uff09 \uff082\uff09[\uff08CUA\uff09 \uff08CUB\uff09] \uff1b\uff083\uff09\uff08A B\uff09 \uff08CUC\uff09
\u4e09\u3001\u89e3\u7b54\u9898
1\uff0em=2\u00d73=6 2.{a } 3.a=-1
4. \u63d0\u793a\uff1a\u4ee4f(1)<0 \u4e14f(2)<0\u89e3\u5f97
5\uff0e\u63d0\u793a\uff1aA={0\uff0c-4}\uff0c\u53c8A B=B\uff0c\u6240\u4ee5B A
\uff08\u2160\uff09B= \u65f6\uff0c 4\uff08a+1\uff092-4(a2-1)<0\uff0c\u5f97a<-1
(\u2161)B={0}\u6216B={-4}\u65f6\uff0c 0 \u5f97a=-1
\uff08\u2162\uff09B={0\uff0c-4}\uff0c \u89e3\u5f97a=1
\u7efc\u4e0a\u6240\u8ff0\u5b9e\u6570a=1 \u6216a -1
6\uff0eU={1\uff0c2\uff0c3\uff0c4\uff0c5} A={1\uff0c4}\u6216A={2\uff0c3} CuA={2,3,5}\u6216{1\uff0c4\uff0c5} B={3\uff0c4}\uff08CUA\uff09 B=\uff081\uff0c3\uff0c4\uff0c5\uff09\uff0c\u53c8 B={3\uff0c4} CUA={1\uff0c4\uff0c5} \u6545A\u53ea\u6709\u7b49\u4e8e\u96c6\u5408{2\uff0c3}
P=-\uff083+4\uff09=-7 q=2\u00d73=6
7\uff0e\u65b9\u7a0bx2-ax-b=0\u7684\u89e3\u96c6\u4e3a{2\uff0c3}\uff0c\u7531\u97e6\u8fbe\u5b9a\u7406a=2+3=5,b=2\u00d73=6,\u4e0d\u7b49\u5f0fbx2-ax+1>0\u5316\u4e3a6x2-5x+1>0 \u89e3\u5f97{x }
8.\u7531A B \u77e5\u65b9\u7a0b\u7ec4
\u5f97x2+(m-1)x=0 \u57280 x \u5185\u6709\u89e3\uff0c \u5373m 3\u6216m -1\u3002
\u82e5 3\uff0c\u5219x1+x2=1-m<0,x1x2=1,\u6240\u4ee5\u65b9\u7a0b\u53ea\u6709\u8d1f\u6839\u3002
\u82e5m -1,x1+x2=1-m>0,x1x2=1,\u6240\u4ee5\u65b9\u7a0b\u6709\u4e24\u6b63\u6839\uff0c\u4e14\u4e24\u6839\u5747\u4e3a1\u6216\u4e24\u6839\u4e00\u4e2a\u5927\u4e8e1\uff0c\u4e00\u4e2a\u5c0f\u4e8e1\uff0c\u5373\u81f3\u5c11\u6709\u4e00\u6839\u5728[0\uff0c2]\u5185\u3002
\u56e0\u6b64{m <m -1}\u3002
\u5982\u56fe:
\u5982\u679c\u6ee1\u610f\u8bf7\u91c7\u7eb3\u6216\u8005\u70b9\u8d5e\uff0c\u8c22\u8c22
f(x)=(sinx+cosx)²-2cos²x-m 在x∈［0，π］上有两个零点x1，x2 求tan(x1+x2)
令f(x)=(sinx+cosx)²-2cos²x-m =0，方程的解就是函数的零点。
分离变量：m=(sinx+cosx)²-2cos²x x∈［0，π］
转化为函数y=m与函数y=(sinx+cosx)²-2cos²x图像在x∈［0，π]上有两个交点，交点的横坐标即为x1，x2
y=(sinx+cosx)²-2cos²x
=1+2sinxcosx--2cos²x （用二倍角公式）
=sin2x-cos2x （用辅助角公式）
=√2sin(2x-π/4)
对称轴为2x-π/4=π/2+2kπ
x=3π/8+kπ
2x-π/4∈[-π/4，3π/4] 所以 sin(2x-π/4)∈[-√2/2，1]
所以1<m≤√2
(x1+x2)/2=3π/8
x1+x2=3π/4
tan(x1+x2)不存在=tan(3π/4)=-1。
供参考。
绛旓細寰堢畝鍗曠殑涓閬楂樹竴鏁板鍗曡皟鎬ч 鏂规硶1锛 瑙ｏ細 y=k/x 璁撅細x1锛渪2 鏈夛細y(x2)-y(x1)=k(1/x2-1/x1) 鍗筹細y(x2)-y(x1)=k(x1-x2)/[(x1)(x2)] 1銆佸綋k锛0鏃讹紝鏈夛細y(x2)-y(x1)锛0 姝ゆ椂锛寉涓哄噺鍑藉紡锛 2銆佸綋k锛0鏃讹紝鏈夛細y(x2)-y(x1)锛0 姝ゆ椂锛寉涓...
绛旓細绗41棰 闃垮皵鍝堟．寮瑰瓙闂Alhazen's Billiard Problem 鍦ㄤ竴涓凡鐭ュ渾鍐,浣滃嚭涓涓叾涓よ叞閫氳繃鍦嗗唴涓や釜宸茬煡鐐圭殑绛夎叞涓夎褰傜42棰 鐢卞叡杞崐寰勪綔妞渾An Ellipse from Conjugate Radii 宸茬煡涓や釜鍏辫江鍗婂緞鐨勫ぇ灏忓拰浣嶇疆,浣滄き鍦嗐 绗43棰 鍦ㄥ钩琛屽洓杈瑰舰鍐呬綔妞渾An Ellipse in a Parallelogram 鍦ㄨ瀹氱殑骞宠鍥涜竟褰㈠唴浣滀竴...
绛旓細瑙ｏ細锛1锛夌敱棰寰梐2a1=a1+a1+a2,a2a2=a1+a2+a1+a2銆傗埓a2²=a1a2+a2.褰揳2=0鏃讹紝a1=0.褰揳2鈮0鏃讹紝寰梐1=a2-1.浜庢槸a2²=4a2-4锛宎2=2.鈭碼1=1.缁间笂锛宎1=a2=0鎴朼1=1锛宎2=2.锛2锛夌敱棰樻剰鐭1=1锛宎2=2.鈭2an=Sn+3.鈭2a(n-1)=S(n-1)+3.涓ゅ紡鐩稿噺寰2锛坅n-a...
绛旓細1 锛1锛夛紝鏈塮锛坸y锛=f锛坸锛+f锛坹锛変护x=y=1锛屽垯f锛1锛=f锛1锛+f锛1锛夛紝鍗砯锛1锛=2f锛1锛夆埓 f锛1锛=0 锛2锛夛紝f锛坸锛夋槸瀹氫箟鍦紙0锛+鈭烇級涓婄殑鍑芥暟 鈭 x锛0锛2-x锛0 鈭 x鈭堬紙0锛2锛夋牴鎹鎰忥紝f锛坸锛+f锛2-x锛=f銆恱锛2-x锛夈戝彟f锛1/3锛=1锛屽垯f锛1/9锛=f锛...
绛旓細锛1锛夎x杞村垎鎴愮殑涓ゆ鍦嗗姬寮ч暱涔嬫瘮涓3锛1 璇存槑鍦嗗績瑙掍负360掳脳1/4=90掳 |b|=锛堟牴鍙2锛塺/2 锛2锛夊渾C鎴獃杞存墍寰楀鸡闀夸负2 璇存槑|a|=鏍瑰彿锛坮²-1锛夛紙3锛夊渾蹇僀鍒扮洿绾縧锛歺-2y=0鐨勮窛绂讳负5鍒嗕箣鏍瑰彿5 鏍规嵁鐐瑰埌绾跨殑璺濈鍏紡锛殀a-2b|/锛堟牴鍙5锛=锛堟牴鍙5锛/5 |a-2b|=1 鐢憋紙1锛...
绛旓細姣忔壒璐叆x鍙帮紝鍒欏叡闇鍒3600/x 鎵癸紝姣忔壒璐圭敤2000x鍏冿紟鐢遍鎰忕煡y=3600/x脳400+k脳2000x锛屽綋x=400鏃讹紝y=43600锛岃В寰梜=1/20 锛2锛夌敱锛1锛夌煡锛寉=3600/x脳400+100x鈮 24000锛堝厓锛夊綋涓斾粎褰 3600/x脳400=100x锛屽嵆x=120鏃剁瓑鍙锋垚绔嬶紟鏁呭彧闇姣忔壒璐叆120鍙帮紝鍙互浣胯祫閲戝鐢紟...
绛旓細绗竴澶╋細缁1/7鐨勶紝绗簩澶╋細缁2/7鐨勶紝鏀跺洖1/7鐨 绗笁澶╋紝缁1/7鐨 绗洓澶╋細缁4/7鐨勶紝鏀跺洖1/7鍜2/7鐨 绗簲澶╋細缁1/7鐨 绗叚澶╋細缁2/7鐨勶紝鏀跺洖1/7鐨 绗竷澶╋細缁1/7鐨 2.闃垮熀绫冲痉鍒嗙墰闂Archimedes'Problema Bovinum 澶槼绁炴湁涓鐗涚兢锛岀敱鐧姐侀粦銆佽姳銆佹鍥涚棰滆壊鐨勫叕銆佹瘝鐗涚粍鎴.鍦...
绛旓細銆楂樹竴鏁板銆戞眰杩欎竴棰樼殑瑙ｇ瓟杩囩▼锛岃阿璋 d鐨勬渶澶у兼槸AB=鏍瑰彿[(1+3)^2+(4-1)^2]=5,鏁呰寖鍥存槸[0,5] 褰撶洿绾縇鍨傜洿浜嶢B鏃,d鏈澶,鍒欐湁K(AB)=(4-1)/(1+3)=3/4 鏁呯洿绾縇鐨勬枩鐜噆=-1/(3/4)=-4/3 鍗崇洿绾縇鐨勬柟绋嬫槸y-1=-4/3(x+3),鍗虫湁y=-4/3x-3 璁剧洿绾縇鏂圭▼鏄痽...
绛旓細鈭迪=4/5<4/3婊¤冻棰鎰忥紝蠅=2>4/3涓嶆弧瓒抽鎰 缁间笂锛氭弧瓒抽鎰忥紝涓斿湪锛0锛屜/8]涓婂崟璋冮掑鐨勫嚱鏁拌В鏋愬紡鍙湁f(x)=2sin(4/5x+蟺/3)(3) 瑙ｆ瀽锛氳鍑芥暟f(x)鍦╗0锛2]涓婃伆鏈変竴涓渶澶у煎拰鏈灏忓 鈭礷(x)=2sin(蠅x+蟺/3)鍗曡皟閫掑噺鍖洪棿锛2k蟺+蟺/2<=蠅x+蟺/3<=2k蟺+3蟺/2==...
绛旓細瑙ｇ瓟濡備笅锛1銆1銆佺敱浜嶴5脳S6+15=0 S5=5 鎵浠5 脳S6+15=0 鎵浠6=-3 a6=S6-S5=-8 鍥犱负绛夊樊鏁板垪Sn=锛坅1+an锛*n/2 鎵浠6=锛坅1+a6锛*3 鍥犳鎺ㄥ嚭a1=7 d=-3 1銆2銆佺敱an鏄椤逛负a1,鍏樊涓篸鐨勭瓑宸暟鍒 鎵浠n=锛坣-1锛*d+a1 Sn=锛坅1+an锛*n/2=锛2a1+锛坣-1锛...
