CMR A=3^2/10^2+3^2/11^2+….+3^2/99^2>4/5

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1. Ta có: `3^2/10^2 = 3^2/10.10 > 3^2/10.11`

   `3^2/11^2 = 3^2/11.11 > 3^2/11.12`

   `………………………………………..`

   `3^2/99^2 = 3^2/99.99 > 3^2/99.100`

   `=> 3^2/10^2 + 3^2/11^2 +….+3^2/99^2 > 3^2/10.11 + 3^2/11.12 + ….+3^2/99.100`

   `=> A > 3^2( 1/10.11 + 1/11.12 +…+1/99.100)`

   `=> A > 3^2( 1/10- 1/11 + 1/11 – 1/12 +…+1/99-1/100)`

   `=> A > 3^2( 1/10 – 1/100)`

   `=> A > 9. 9/100`

   `=> A > 81 /100`

   Vì `81/100 > 4/5`

   `=> A> 4/5`

   Vậy `A> 4/5`

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2. Bài làm :

   Ta có :

   \(\left\{ \begin{array}{l}\dfrac{3^2}{10^2} > \dfrac{3^2}{10 . 11}\\ \dfrac{3^2}{11^2} > \dfrac{3^2}{11 . 12} \\………..\\ \dfrac{3^2}{99^2} > \dfrac{3^2}{99 . 100}\end{array} \right.\)

   `-> A > 3^2/(10 . 11) + 3^2/(11 . 12) + … + 3^2/(99 . 100)`

   `-> A > 3^2[1/(10 . 11) + 1/(11 . 12) + …. + 1/(99 . 100)]`

   `-> A > 3^2 [1/10 – 1/11 + 1/11 – 1/12 + … + 1/99 – 1/100]`

   `-> A > 3^2 [1/10 + (- 1/11 + 1/11 – 1/12 + … + 1/99) – 1/100]`

   `-> A > 3^2 [1/10 – 1/100]`

   `->A > 3^2 . 9/100`

   `-> A > 9 . 9/100`

   `-> A > 81/100`

   Ta thấy `81/100 > 4/5`

   `-> A > 4/5`

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