cho C= 1/101 + 1/102 + 1/103+…+ 1/200 cmr c >5/8

cho C= 1/101 + 1/102 + 1/103+…+ 1/200
cmr c >5/8

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  1. Đáp án:

    `↓↓` 

    Giải thích các bước giải:

    `C=1/101+1/102+….+1/200`

    `=(1/101+1/102 +….+1/120) + (1/121+1/122+….+1/150) + (1/151+1/152+….+1/180) + (1/181+1/182+….+1/200)` 

    `C>20 . 1/120 + 30 . 1/150 + 30 . 1/180 + 20 . 1/200`

    `> 1/ 6 + 1/ 5 + 1/ 6 + 1/ 10`

    `> 19/30 = 76/120 `

    `> 75/120 + 1/120`

    `> 5/ 8 +1/120>5/8`

    `=> đpcm`

     

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  2. `C=1/101+1/102+…+1/200`

    `C=(1/101+1/102 +…+1/120)+(1/121+1/122+..+1/150)+(1/151+1/152+…+1/180)+(1/181+1/182+…+1/200)`

    `C>20. 1/120+30. 1/150+30. 1/180+20. 1/200`

    `C>1/ 6 + 1/ 5 + 1/ 6 + 1/ 10`

    `C>(1/ 6 +1/ 6)+(1/ 5 +1/10)`

    `C>1/ 3 + 3/ 10`

    `C>19/30=76/120`

    `C>75/120+1/120`

    `C>5/ 8 +1/120`

    Vậy `C>5/ 8`

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