chứng tỏ rằng 200-(3+2/3+2/4+2/5+…+2/100)/ 1/2+2/3+3/4+…+99/100=2

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1. Em tham khảo nhé:

   $\text{Xét 200-(3+$\dfrac{2}{3}$+$\dfrac{2}{4}$+….+$\dfrac{2}{100}$)}$

   $\text{=200-2-(1+$\dfrac{2}{3}$+…+$\dfrac{2}{100}$)}$

   $\text{=198-(1+$\dfrac{2}{3}$+…+$\dfrac{2}{100}$)}$

   $\text{=2[99-($\dfrac{1}{2}$- $\dfrac{1}{3}$+….+ $\dfrac{1}{100}$ ) (1)}$

   $\text{Xét $\dfrac{1}{2}$+$\dfrac{2}{3}$+..+ $\dfrac{99}{100}$}$

   $\text{⇒Rút 1 ra (Có 99 số)}$

   $\text{⇒99-(  $\dfrac{1}{2}$+ $\dfrac{1}{3}$+..+$\dfrac{1}{100}$) (2)}$

   $\text{Từ 1,2⇒A=2}$

    

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2. Ta có:

       $\dfrac{200-\bigg(3+\dfrac{2}{3}+…+\dfrac{2}{100}\bigg)}{\dfrac{1}{2}+\dfrac{2}{3}+…+\dfrac{99}{100}}$

   $=\dfrac{200-2-\bigg(\dfrac{2}{2}+\dfrac{2}{3}+…+\dfrac{2}{100}\bigg)}{1-\dfrac{1}{2}+1-\dfrac{2}{3}+…+1-\dfrac{1}{100}}$

   $=\dfrac{198-\bigg(\dfrac{2}{2}+\dfrac{2}{3}+…+\dfrac{2}{100}\bigg)}{\bigg(1+1+…+1)-(\dfrac{1}{2}+\dfrac{1}{3}+…+\dfrac{1}{100}\bigg)}$

   $=\dfrac{2.\bigg[99-\bigg(\dfrac{1}{2}+\dfrac{1}{3}+…+\dfrac{1}{100}\bigg)\bigg]}{99-\bigg(\dfrac{1}{2}+\dfrac{1}{3}+…+\dfrac{1}{100}\bigg)}$

   $=2$

   Vậy… (đpcm)

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