北师大版高中（必修一）数学，请教几个题的做题步骤，详细点最好，在线等
\u5317\u5e08\u5927\u7248\u9ad8\u4e2d\uff08\u5fc5\u4fee\u4e00\uff09\u6570\u5b66\uff0c\u8bf7\u6559\u51e0\u4e2a\u9898\u7684\u505a\u9898\u6b65\u9aa4\uff0c\u8be6\u7ec6\u70b9\u6700\u597d\uff0c\u5728\u7ebf\u7b49\u9996\u5148X<0; g(x)=-x²<0
f\uff08g\uff08x\uff09\uff09=f\uff08-x²\uff09=-(-x²)=x²
1\u3001\u5df2\u77e5\u975e\u96f6\u5411\u91cfAB\u4e0eAC\u6ee1\u8db3[\uff08AB/\uff5cAB\uff5c\uff09+ (AC/\uff5cAC\uff5c)]•BC=0\uff0c
\u4e14\uff08AB/\uff5cAB\uff5c)•(AC/\uff5cAC\uff5c) =½ \uff0c\u5224\u65ad\u4e09\u89d2\u5f62ABC\u7684\u5f62\u72b6\u3002
(\u539f\u9898\u5199\u6f0f\u534a\u4e2a\u4e2d\u62ec\u53f7)(AB/\u2502AB\u2502\u8868\u4e0e\u5411\u91cfAB\u540c\u5411\u7684\u5355\u4f4d\u5411\u91cf,\u5176\u6a21=1.\u5176\u4f59\u7c7b\u4f3c)
\u89e3:\uff08AB/\uff5cAB\uff5c)•(AC/\uff5cAC\uff5c)=1\u00d71\u00d7cosA =½ ,\u6545A=60\u00b0
[\uff08AB/\uff5cAB\uff5c\uff09+ (AC/\uff5cAC\uff5c)]•BC=\u2502(AB/\uff5cAB\uff5c\uff09+ (AC/\uff5cAC\uff5c)\u2502\u2502BC\u2502cos(A/2+C)=0
\u5f97cos(A/2+C)=0\u6545A/2+C=90\u00b0,\u2234C=90\u00b0-60\u00b0/2=60\u00b0,
\u25b3ABC\u662f\u7b49\u8fb9\u25b3.
2\u3001\u5728\u56db\u8fb9\u5f62ABCD\u4e2d\uff0cBD\u662f\u5b83\u7684\u4e00\u6761\u5bf9\u89d2\u7ebf\uff0c\u4e14 BC=\u03bb(AD)\uff08\u03bb\u2208R\uff09\uff0c\uff5cAB\uff5c=\uff5cAD\uff5c=2\uff0c
\uff5cCB-CD\uff5c=2\u221a3
\uff081\uff09\u3001\u82e5\u4e09\u89d2\u5f62BCD\u4e3a\u76f4\u89d2\u4e09\u89d2\u5f62\uff0c\u6c42\u03bb\u7684\u503c
\uff082\uff09\u3001\u5728\uff081\uff09\u7684\u6761\u4ef6\u4e0b\uff0c\u6c42 CB•BA
\u89e3:(1) \u2502CB-CD\u2502=\u2502DB\u2502=2\u221a3
\u5728\u25b3ABD\u4e2d,\u2502DB\u2502²=\u2502AD\u2502²+\u2502AB\u2502²-2\u2502AD\u2502\u2502AB\u2502cosA
\u5373\u670912=4+4-8cosA,\u6545cosA=-1/2, \u2234A=120\u00b0, \u2220ABD=\u2220ADB=30\u00b0
BC=\u03bb(AD),\u6545BC\u2016AD,\u4e14\u2502BC\u2502=\u03bb\u2502AD\u2502=2\u03bb
\u2220DBC=\u2220ABC-\u2220ABD=60\u00b0-30\u00b0=30\u00b0
\u2220C=90\u00b0,\u6545\u2502BC\u2502=2(\u221a3)cos30\u00b0=3=2\u03bb, \u2234\u03bb=3/2
(2)CB•BA=\u2502CB\u2502\u2502BA\u2502cos120\u00b0=3\u00d72\u00d7(-1/2)=-3
3\u3001\u4ee5\u539f\u70b9\u548c\u70b9A\uff085\uff0c2\uff09\u4e3a\u9876\u70b9\u4f5c\u7b49\u8170\u76f4\u89d2\u4e09\u89d2\u5f62OAB\uff0c\u4f7f\u2220B=90\u00b0\uff0c
\u6c42\u70b9B\u548c\u5411\u91cfAB \u7684\u5750\u6807\u3002
\u89e3:\u2502OA\u2502=\u221a29, OA\u4e2d\u70b9M(5/2, 1), \u4ee5M\u4e3a\u5706\u5fc3,\u4ee5\u2502OA\u2502/2=(\u221a29)/2\u4e3a\u534a\u5f84\u4f5c\u56edM:
M: (x-5/2)²+(y-1)²=29/4
\u8fc7M\u4f5cOA\u7684\u5782\u76f4\u7ebf: y=-(5/2)(x-5/2)+1=-(5/2)x+29/4,\u4ee3\u5165\u56edM\u7684\u65b9\u7a0b,\u5316\u7b80\u5f97
4x²-20x+21=(2x-7)(2x-3)=0
\u89e3\u5f97x₁=3.5, x₂=1.5.
\u4e8e\u662fy₁=-1.5, y₂=3.5
\u5373B₁(3.5, -1.5); B₂(1.5, 3.5)
\u5411\u91cfAB₁=-1.5i - 3.5j
\u5411\u91cfAB₁=-3.5i +1.5j
4\u3001\u5df2\u77e5O\u3001A\u3001B\u4e09\u70b9\u7684\u5750\u6807\u5206\u522b\u4e3aO(0\uff0c0)\uff0cA(3\uff0c0)\uff0cB\uff080\uff0c3\uff09\uff0c\u70b9P\u5728\u7ebf\u6bb5AB\u4e0a\uff0c
\u4e14 \u5411\u91cfAP=t\u500d\u5411\u91cfAB\uff0c\uff080\u2264t\u22641\uff09\uff0c\u5219\u5411\u91cfOA•\u5411\u91cfOP\u7684\u6700\u5927\u503c\u4e3a\uff3f\u3002
\u89e3:OA•OP=\u2502OA\u2502\u2502OP\u2502cos\u2220AOP\u2264\u2502OA\u2502²=9
\u5f53t=0\u5373P\u70b9\u4e0eA\u70b9\u91cd\u5408\u65f6OA•OP\u83b7\u5f97\u6700\u5927\u503c9.
5\u3001\u4e0e\u5411\u91cfa=\uff087/2\uff0c½\uff09\u548c\u5411\u91cfb=\uff08½\uff0c7/2\uff09\u7684\u5939\u89d2\u76f8\u7b49\uff0c\u4e14\u6a21\u4e3a1\u7684\u5411\u91cf\u7684\u5750\u6807\u662f\uff3f\u3002
\u89e3:\u4e0e\u5411\u91cfa\u540c\u5411\u7684\u5355\u4f4d\u5411\u91cfa\u00b0=a/\u2502a\u2502=a/(25/2)=2a/25
\u4e0e\u5411\u91cfb\u540c\u5411\u7684\u5355\u4f4d\u5411\u91cfb\u00b0=b/\u2502b\u2502=b/(25/2)=2b/25
a\u00b0\u4e0eb\u00b0\u7684\u548c\u5411\u91cfc=a\u00b0+b\u00b0=(2/25)(a+b)
\u5411\u91cfc\u5e73\u65b9\u5411\u91cfa\u548cb\u7684\u67b6\u89d2.
\u4e0e\u5411\u91cfc\u540c\u5411\u7684\u5355\u4f4d\u5411\u91cfc\u00b0=c/\u2502c\u2502=c/(2/25)\u221a2=(a+b)/\u221a2=(\u221a2)a/2+(\u221a2)b/2
\u6545\u4e0e\u5411\u91cfa,b\u5939\u89d2\u76f8\u7b49\u7684\u5355\u4f4d\u5411\u91cfc\u00b0\u7684\u5750\u6807\u4e3a(\u221a2/2, \u221a2/2)
6\u3001\u5df2\u77e5\u4e09\u70b9A(1\uff0c2)\uff0cB\uff083\uff0c1\uff09\uff0cC\uff08-1\uff0c0\uff09\uff0c\u8bd5\u56de\u7b54\u4e0b\u5217\u95ee\u9898\uff1a
\uff081\uff09\u3001\u7528\u5750\u6807\u8868\u793a\u5411\u91cfAB\uff0c\u5e76\u6c42\u5b83\u7684\u6a21\uff1b
\uff082\uff09\u3001\u6c42\u4f7f\u5411\u91cfAB=\u5411\u91cfCD\u7684\u70b9D\u7684 \u5750\u6807\uff1b
\uff083\uff09\u3001\u8bbe\u5411\u91cfAB\u548c\u5411\u91cfAC\u7684\u5939\u89d2\u4e3a\u03b8\uff0c\u6c42cos\u03b8 \u7684\u503c\uff1b
\uff084\uff09\u3001\u6c42\u5e73\u884c\u56db\u8fb9\u5f62ABCD\u7684\u9762\u79ef\u3002
\u89e3:(1)AB=(3-1, 1-2)=(2, -1), \u2502AB\u2502=\u221a[2²+(-1)²]=\u221a5
(2)\u8bbeD(x, y),\u5219CD=(x+1, y-0)=(2, -1)
\u5176\u4e2dx+1=2, x=1, y=-1,\u6545D(1,-1)
(3)AC=(-2, -2)
AB\u6240\u5728\u76f4\u7ebf\u7684\u659c\u7387k₂=-1/2; AC\u6240\u5728\u76f4\u7ebf\u7684\u659c\u7387k₁=1
\u6545\u4eceAC\u5230AB\u7684\u5939\u89d2\u03b8\u7684\u6b63\u5207tan\u03b8=(k₂-k₁)/(1+k₁k₂)=(-1/2-1)/(1-1/2)=-3
\u4e8e\u662f\u5f97cos\u03b8=-1/\u221a(1+tan²\u03b8)=-1/\u221a10, sin\u03b8=\u221a(1-1/10)=3/\u221a10,
(4)\u5e73\u5f62\u56db\u8fb9ABCD\u7684\u9762\u79efS=\u2502AB\u2502\u2502AC\u2502sin\u03b8=(\u221a5)\u00d7(\u221a8)\u00d7(3/\u221a10)=6
7\u3001\u5e73\u9762\u5185\u5411\u91cfOA=\uff081\uff0c7\uff09\uff0c\u5411\u91cfOB=\uff085\uff0c1\uff09\uff0c\u5411\u91cfOP=\uff082\uff0c1\uff09\uff0c\u70b9Q\u4e3a\u76f4\u7ebfOP
\u4e0a\u7684\u4e00\u4e2a\u52a8\u70b9\u3002
\uff081\uff09\u3001\u5f53\u5411\u91cfQA•\u5411\u91cfQB\u53d6\u6700\u5c0f\u503c\u65f6\uff0c\u6c42\u5411\u91cfOQ \u7684\u5750\u6807\uff1b
\uff082\uff09\u3001\u5f53\u70b9Q\u6ee1\u8db3\uff081\uff09\u7684\u6761\u4ef6\u548c\u7ed3\u8bba\u65f6\uff0c\u6c42cos\u2220AQB \u7684\u503c\u3002
\u89e3:(1)Q\u5728OP\u4e0a,\u6545\u53ef\u8bbeQ\u7684\u5750\u6807\u4e3a(2y,y),\u5176\u4e2d0\u2264y\u22641.
QB=(5-2y, 1-y), QA=(1-2y, 7-y)
QA•QB=(5-2y)(1-2y)+(1-y)(7-y)=5y²-20y+12=5(y-2)²-8
\u5f53y=1\u65f6QA•QB\u83b7\u5f97\u6700\u5c0f\u503c(-3)
(2)\u6b64\u65f6Q(2, 1), QB=(3, 0); QA=(-1, 6)
cos\u2220AQB=QA•QB/\u2502QA\u2502\u2502QB\u2502=-3/(3\u221a37)=-1/\u221a37.
8\u3001\u5df2\u77e5\u5411\u91cfa\uff0cb\u4e3a\u975e\u96f6\u5411\u91cf\uff0c\u5f53 \u5411\u91cfa+t\u500d\u5411\u91cfb \uff08t\u2208R\uff09\u7684\u6a21\u53d6\u6700\u5c0f\u503c\u65f6\uff1a
\uff081\uff09\u3001\u6c42t\u7684\u503c\uff1b
\uff082\uff09\u3001\u6c42\u8bc1\uff1a\u5411\u91cfb \u4e0e \u5411\u91cfa+t\u500d\u5411\u91cfb \u5782\u76f4\u3002
\u89e3:(1)\u4e3a\u4f7f\u95ee\u9898\u7b80\u5316,\u53d6a,b\u7684\u4ea4\u70b9O\u4f5c\u5750\u6807\u539f\u70b9,\u5411\u91cfb\u5728x\u8f74\u4e0a\u4e14\u4e0ex\u8f74\u540c\u5411,a\u5728\u7b2c\u4e00\u8c61\u9650
\u5185,a\u4e8eb\u7684\u5939\u89d2\u4e3a\u9510\u89d2.\u4e8e\u662f\u53ef\u8bbea=(m,n), b=(k,0), (m\uff1e0, n\uff1e0, k\uff1e0)
a+tb=(m+tk, n)
\u2502a+tb\u2502=\u221a[(m+kt)²+n²]=\u221a(k²t²+2mkt+m²+n²)=\u221a[k²(t+m/k)²+n²]\u2265n
\u5f53t=-m/k\u65f6\u7b49\u53f7\u6210\u7acb,\u6b64\u65f6\u2502a+tb\u2502min=n, a+tb=(0, n)
(2)b•(a+tb)=k\u00d70+0\u00d7n=0,\u53c8b•(a+tb)=\u2502b\u2502\u2502a+tb\u2502cos\u03b8=0
\u5176\u4e2d\u03b8\u4e3aa\u4e0ea+tb\u7684\u5939\u89d2,\u2502b\u2502\u22600, \u2502a+tb\u2502\u22600,\u6545\u5fc5\u6709cos\u03b8=0,\u5373\u03b8=90\u00b0
\u4e5f\u5c31\u662fb\u22a5(a+tb), \u6545\u8bc1.
9\u3001\u5df2\u77e5AD \u3001BE\u3001CF\u662f\u4e09\u89d2\u5f62ABC\u7684\u4e09\u6761\u9ad8\uff0c\u6c42\u8bc1\uff1aAD \u3001BE\u3001 CF\u76f8\u4ea4\u4e8e\u4e00\u70b9\u3002
\u89e3:\u6b64\u9898\u7528\u77e2\u91cf\u597d\u50cf\u4e0d\u597d\u8bc1.\u4f60\u770b\u770b\u521d\u7b49\u51e0\u4f55\u5427.\u4e0b\u56de\u522b\u4e00\u6b21\u63d0\u8fd9\u4e48\u591a\u95ee\u9898,\u592a\u8d39\u65f6\u95f4\u4e86!
2.f(x-1)=x^2=(x-1)^2+2(x-1)+1得，f(x)=x^2+2x+1
3.设f(x)=ax+b 则f(f(x))=f(ax+b)=a(ax+b)+b=a^2x+ab+b=9x+4
则有，a^2=9,ab+b=4,解得a=3，b=1，或a=-3，b=-2
f(x)=3x+1或f(x)=-3x-2
1.f(x-1)=(x-1)^2
2.设t=x-1,f(t)=(t+1)^2,f(x)=(x+1)^2,
3. 设f(x)=kx+b,f（f（x））=k(kx+b)+b=9x+4,k^2=9;kb+b=4
k=3.b=1,,,,,,k=-3,b=-2
分别是换元法和待定系数法
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绛旓細涓锛屽鍑芥暟锛欶(x)=-F(-x)鍋跺嚱鏁帮細F(x)=F(-x)浜岋紝 鍛ㄦ湡鍑芥暟锛欶(x+T)=F(x) T 鏄懆鏈 涓夛紝鍑芥暟鐨勮繍绠楋細F(x)锛孏(x)鐨勫畾涔夊煙涓篋1,D2,D=D1nD2涓嶇瓑浜庣┖闆 鍜 F+G 锛團+G)(x)=F(x)+G(x) X灞炰簬D (宸篃鏄姝わ級绉 锛團*G)(x)=F(x)*G(x) X灞炰簬...
绛旓細棣栧厛X<0; g(x)=-x²<0 f锛坓锛坸锛夛級=f锛-x²锛=-(-x²)=x²
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