离散数学 R1={(a,b)|a≤b} R2={(a,b)|a=b或a=-b} 大学高数 设(a×b)·c=2,则{(a+b)×(b+c)}...

\u8bbe\u6709\u5173\u7cfb\u6a21\u5f0fR (A\uff0cB\uff0cC\uff0cD\uff0cE)\uff0c\u5176\u4e0a\u7684\u51fd\u6570\u4f9d\u8d56\u96c6\uff1aF={A\u2192BC\uff0cCD\u2192E\uff0cB\u2192D\uff0cE\u2192A}\uff081\uff09\u8ba1\u7b97B+\u3002\uff082\uff09\u6c42

\u9996\u5148\u628a\u51fd\u6570\u4f9d\u8d56\u5206\u6210\u5355\u4e00\u7684\u51fd\u6570\u4f9d\u8d56\uff0cF=(A\u2192C\uff0cC\u2192A\uff0cB\u2192A\uff0cB->C\uff0cD\u2192A\uff0cD->C), \u56e0\u4e3aD\u2192A\uff0cA\u2192C\uff0c\u6240\u4ee5D->C\u5197\u4f59\u4e86\uff0c\u6240\u4ee5\u5220\u53bb\u3002\u540c\u6837\uff0cB->C\uff0cC\u2192A\uff0c\u6240\u4ee5B\u2192A\u5220\u53bb\u3002 Fmin=(A\u2192C\uff0cC\u2192A\uff0cD\u2192A\uff0cB->C)

\uff5b\uff08a+b\uff09\u00d7(b+c)\uff5d\u00b7(c+a)=4\u3002
\u5206\u6790\u8fc7\u7a0b\u5982\u4e0b\uff1a
\uff5b\uff08a+b\uff09\u00d7(b+c)\uff5d\u00b7(c+a)={a\u00d7b+b\u00d7b+a\u00d7c+bxc}\u00b7(c+a)=(a\u00d7b+0+a\u00d7c+bxc)(c+a) [\u6ce8\u610f\uff1ab\u00d7b=0]
=(a\u00d7b)\u00b7c+ ( b\u00d7c )\u00b7a [\u6ce8\u610f\uff1a(a\u00d7c)\u00b7c=0,\u3010\u2235a\u00d7c\u22a5c\u3011\uff0c\u540c\u68370=(b\u00d7c)\u00b7c=(a\u00d7b)\u00b7a=(a\u00d7c)\u00b7a]
=2(a\u00d7b)\u00b7c=2\u00d72=4\u3002

\u6269\u5c55\u8d44\u6599\uff1a
\u5411\u91cf\u79ef|c|=|a\u00d7b|=|a||b|sin\uff0c\u5373c\u7684\u957f\u5ea6\u5728\u6570\u503c\u4e0a\u7b49\u4e8e\u4ee5a\uff0cb\uff0c\u5939\u89d2\u4e3a\u03b8\u7ec4\u6210\u7684\u5e73\u884c\u56db\u8fb9\u5f62\u7684\u9762\u79ef\u3002\u800cc\u7684\u65b9\u5411\u5782\u76f4\u4e8ea\u4e0eb\u6240\u51b3\u5b9a\u7684\u5e73\u9762\uff0cc\u7684\u6307\u5411\u6309\u53f3\u624b\u5b9a\u5219\u4ecea\u8f6c\u5411b\u6765\u786e\u5b9a\u3002*\u8fd0\u7b97\u7ed3\u679cc\u662f\u4e00\u4e2a\u4f2a\u5411\u91cf\u3002\u8fd9\u662f\u56e0\u4e3a\u5728\u4e0d\u540c\u7684\u5750\u6807\u7cfb\u4e2dc\u53ef\u80fd\u4e0d\u540c\u3002
1\u3001\u53cd\u4ea4\u6362\u5f8b\uff1aa\u00d7b=-b\u00d7a
2\u3001\u52a0\u6cd5\u7684\u5206\u914d\u5f8b\uff1aa\u00d7\uff08b+c\uff09=a\u00d7b+a\u00d7c\u3002
3\u3001\u4e0e\u6807\u91cf\u4e58\u6cd5\u517c\u5bb9\uff1a\uff08ra\uff09\u00d7b=a\u00d7\uff08rb\uff09=r\uff08a\u00d7b\uff09\u3002
4\u3001\u4e0d\u6ee1\u8db3\u7ed3\u5408\u5f8b\uff0c\u4f46\u6ee1\u8db3\u96c5\u53ef\u6bd4\u6052\u7b49\u5f0f\uff1aa\u00d7\uff08b\u00d7c\uff09+b\u00d7\uff08c\u00d7a\uff09+c\u00d7\uff08a\u00d7b\uff09=0\u3002
5\u3001\u5206\u914d\u5f8b\uff0c\u7ebf\u6027\u6027\u548c\u96c5\u53ef\u6bd4\u6052\u7b49\u5f0f\u522b\u8868\u660e\uff1a\u5177\u6709\u5411\u91cf\u52a0\u6cd5\u548c\u53c9\u79ef\u7684R3\u6784\u6210\u4e86\u4e00\u4e2a\u674e\u4ee3\u6570\u3002
6\u3001\u4e24\u4e2a\u975e\u96f6\u5411\u91cfa\u548cb\u5e73\u884c\uff0c\u5f53\u4e14\u4ec5\u5f53a\u00d7b=0\u3002
\u53c2\u8003\u8d44\u6599\uff1a\u767e\u5ea6\u767e\u79d1\u2014\u2014\u5411\u91cf\u79ef

利用配方法。ax^2+bx+c=a[x^2+(b/a)x]+c=a{x^2+(b/a)x+[b/(2a)]^2}-a[b/(2a)]^2+c=a[x+b/(2a)]^2-b^2/(4a)+4ac/(4a)=a[x+b/(2a)]^2-(b^2-4ac)/(4a)=a[x+b/(2a)]^2-a[√(b^2-4ac)]^2/(2a)^2=a{[x+b/(2a)]^2-[√(b^2-4ac)/(2a)]^2}=a[x+b/(2a)-√(b^2-4ac)/(2a)][x+b/(2a)+√(b^2-4ac)/(2a)]=a{x-[-b+√(b^2-4ac)]/(2a)}{x-[-b-√(b^2-4ac)]/(2a)}=a(x-r1)(x-r2)。离散数学 设R1和R2是A上的关系,证明 20 s(R1 U R2) = s(R1) U s(R2) r(R1 U R2) = r(

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