已知1/2+1/5+1/8+1/11+1/20+1/41+1/110+1/1640=1求:-1/2-1/5-1/8+1/11-1/20-1/41+1/110+1/1640的值.

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已知1/2+1/5+1/8+1/11+1/20+1/41+1/110+1/1640=1求:-1/2-1/5-1/8+1/11-1/20-1/41+1/110+1/1640的值.
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已知1/2+1/5+1/8+1/11+1/20+1/41+1/110+1/1640=1求:-1/2-1/5-1/8+1/11-1/20-1/41+1/110+1/1640的值.
已知1/2+1/5+1/8+1/11+1/20+1/41+1/110+1/1640=1
求:-1/2-1/5-1/8+1/11-1/20-1/41+1/110+1/1640的值.

已知1/2+1/5+1/8+1/11+1/20+1/41+1/110+1/1640=1求:-1/2-1/5-1/8+1/11-1/20-1/41+1/110+1/1640的值.
1/2+1/5+1/8+1/11+1/20+1/41+1/110+1/1640=1,
则:-1/2-1/5-1/8+1/11-1/20-1/41+1/110+1/1640
=-1+2*[(1/11)+(1/110)+(1/1640)]
=-1+2*(165/1640)
=-131/164

令:-1/2-1/5-1/8+1/11-1/20-1/41+1/110+1/1640=X,
两式相减则有;1+2/5+1/4+1/10+2/41=1-X
则X=-(2/5+1/4+1/10+2/41)=?
似乎还比比较麻烦,应该有更简单的办法.