有理数的乘除法的问题

\u6709\u7406\u6570\u7684\u4e58\u9664\u6cd5\u600e\u4e48\u7b97\uff1f\uff1f

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\uff081\uff09\u5982\u679c\u4e00\u4e2a\u6570\u9664\u4ee5\u5b83\u7684\u5012\u6570\uff0c\u5546\u662f1\uff0c\u90a3\u4e48\u8fd9\u4e2a\u6570\u662f\uff08D\uff09 A\uff0e1 B\uff0e2 C\uff0e-1 D\uff0e\u00b11 \uff082\uff09\u82e5\u4e24\u4e2a\u6709\u7406\u6570\u7684\u5546\u662f\u8d1f\u6570\uff0c\u90a3\u4e48\u8fd9\u4e24\u4e2a\u6570\u4e00\u5b9a\u662f\uff08D\uff09 A\uff0e\u90fd\u662f\u6b63\u6570 B\uff0e\u90fd\u662f\u8d1f\u6570 C\uff0e\u7b26\u53f7\u76f8\u540c D\uff0e\u7b26\u53f7\u4e0d\u540c \uff083\uff09 =-1\uff0c\u5219a\u4e3a \uff08B\uff09 A\uff0e\u6b63\u6570 B\uff0e\u8d1f\u6570 C\uff0e\u975e\u6b63\u6570 D\uff0e\u975e\u8d1f\u6570 \uff084\uff09\u82e5a+b0\uff0c\u5219\u4e0b\u5217\u6210\u7acb\u7684\u662f\uff08B\uff09 A\uff0ea>0\uff0cb>0 B\uff0ea0\uff0cb0 2\uff0e\u8ba1\u7b97\u9898 \uff081\uff09\uff08-2 \uff09\u00f7\uff08\uff0d \uff09\u3000=6\u3000 \uff082\uff093.5\u00f7 \u00f7\uff08-1 \uff09\u3000=\uff0d \u3000 \uff083\uff09\uff0d \u00f7\uff08-7\uff09\u00f7\uff08- \uff09 =- \uff084\uff09\uff08-1\uff09\u00f7\uff08+ \uff09\u00f7\uff08- \uff09 = \u63d0\u5347\u80fd\u529b 3\uff0e\u586b\u7a7a\u9898 \uff081\uff09\u82e5a\u3001b\u662f\u4e92\u4e3a\u5012\u6570\uff0c\u52193ab= 3 \uff0e \uff082\uff09\u76f8\u53cd\u6570\u662f\u5b83\u672c\u8eab\u7684\u6570\u6709\u30000\u3000\uff0c\u7edd\u5bf9\u503c\u7b49\u4e8e\u5b83\u672c\u8eab\u7684\u6570\u662f\u3000\u975e\u8d1f\u6570\u3000\uff0c\u5012\u6570\u7b49\u4e8e\u5b83\u672c\u8eab\u7684\u6570\u662f\u30001\uff0c-1\u3000\uff0e \uff083\uff09\u82e5\u30000\uff0e\uff08\u586b\u201c\uff09\u201d\u3001\u201c\u3008\u201d\u3009 \uff084\uff09\u5f53\u3000x=2\u3000\u65f6\uff0c\u4ee3\u6570\u5f0f\u6ca1\u6709\u610f\u4e49\uff0e 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\u8d1f\u6570\u8d1f\u6570\u8d1f\u6570

1.1/30,1/n*(n+1)
2.(-30)*(1/2-2/3+4/5)=-30*1/2+30*2/3-30*4/5=-15+20-24=-19
3.-1.2*4+1.8+1.6-2.8=-4.2
亏损4.2万元
4.-1/n*1/(n+1)=-1/n+1/(n+1)
原式=-1+1/2-1/2+1/3-1/3+1/4-.....+1/2008-1/2008+1/2009
=-1+1/2009
=-2008/2009

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