chứng minh rằng :4/3<1/11+1/12+1/13+....+1/70<2.5 mọi ng giúp giùm ạ

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1. Em tham khảo

   D=$\frac{1}{11}$+$\frac{1}{12}$+….+ $\frac{1}{60}$ 

   ⇒D=S1+S2+S3=($\frac{1}{11}$+…+ $\frac{1}{20}$)+( $\frac{1}{21}$+..+$\frac{1}{30}$)+(  $\frac{1}{31}$+..+ $\frac{1}{60}$ 

   Xét ($\frac{1}{20}$+..+$\frac{1}{20}$)< S1<($\frac{1}{10}$+…+$\frac{1}{10}$)

   =$\frac{1}{2}$<S1<1 (1)

   Xét ( $\frac{1}{30}$ +..+$\frac{1}{30}$)<S2<( $\frac{1}{20}$+..+$\frac{1}{20}$ 

   = $\frac{1}{3}$<S2< $\frac{1}{2}$  (2)

   Xét ($\frac{1}{60}$ +..+$\frac{1}{60}$ )<S3<($\frac{1}{30}$+..+$\frac{1}{30}$)

   =$\frac{1}{2}$<S3<1 (3)

   Từ 1,2,3⇒$\frac{1}{2}$+$\frac{1}{2}$+$\frac{1}{3}$<D<1+$\frac{1}{2}$+1

   ⇒$\frac{4}{3}$<D<$\frac{5}{2}$=2,5

   Vậy đpcm

   Xin câu trả lời hay nhất

   Trả lời

2. Giải thích các bước giải:

   $A=\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+…+\dfrac{1}{70}\\
   =\left ( \dfrac{1}{11}+\dfrac{1}{12}+…+\dfrac{1}{20} \right )+\left ( \dfrac{1}{21}+…+\dfrac{1}{30} \right )+\left ( \dfrac{1}{31}+…+\dfrac{1}{60} \right )+…+\dfrac{1}{70} $
   Ta có
   $\dfrac{1}{10}+…+\dfrac{1}{10}=1>\dfrac{1}{11}+\dfrac{1}{12}+…\dfrac{1}{20}
   >\dfrac{1}{20}+…+\dfrac{1}{20}=\dfrac{10}{20}=\dfrac{1}{2}\\
   \dfrac{1}{20}+\dfrac{1}{20}=\dfrac{10}{20}=\dfrac{1}{2}>\dfrac{1}{21}+\dfrac{1}{22}+…+\dfrac{1}{30}>\dfrac{1}{30}+\dfrac{1}{30}+…+\dfrac{1}{30}=\dfrac{10}{30}=\dfrac{1}{3}\\
   \dfrac{1}{30}+…+\dfrac{1}{30}=\dfrac{30}{30}=1>\dfrac{1}{31}+\dfrac{1}{32}+…+\dfrac{1}{60}>\dfrac{1}{60}+\dfrac{1}{60}+…+\dfrac{1}{60}=\dfrac{30}{60}=\dfrac{1}{2}\\
   \Rightarrow 1+1+\dfrac{1}{2}=\dfrac{5}{2}>A=\left ( \dfrac{1}{11}+\dfrac{1}{12}+…+\dfrac{1}{20} \right )+\left ( \dfrac{1}{21}+…+\dfrac{1}{30} \right )+\left ( \dfrac{1}{31}+…+\dfrac{1}{60} \right )+…+\dfrac{1}{70} >\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{4}{3}\\
   \Rightarrow \dfrac{5}{2}>A>\dfrac{4}{3}$

   Trả lời