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27 tháng 6 2016

Toán lớp 7

25 tháng 6 2016

\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)

\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)

\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\)

\(A=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)

\(A=\left(\frac{1}{26}+\frac{1}{27}+...+\frac{1}{30}\right)+\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\right)\)

\(A< \left(\frac{1}{25}+\frac{1}{25}+...+\frac{1}{25}\right)+\left(\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}\right)+\left(\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}\right)\)

                5 phân số 1/25                                 10 phân số 1/30                                 10 phân số 1/40

\(A< 5.\frac{1}{25}+10.\frac{1}{30}+10.\frac{1}{40}\)

\(A< \frac{1}{5}+\frac{1}{3}+\frac{1}{4}\)

\(A< \frac{1}{4}+\frac{1}{3}+\frac{1}{4}\)

\(A< \frac{1}{2}+\frac{1}{3}\)

\(A< \frac{5}{6}\)

25 tháng 6 2016

\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=\frac{5}{6}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}\)

\(=\frac{5}{6}+\left(\frac{1}{5}-\frac{1}{4}\right)+\left(\frac{1}{7}-\frac{1}{6}\right)+...+\left(\frac{1}{49}-\frac{1}{48}\right)-\frac{1}{50}\)

\(\frac{1}{5}-\frac{1}{4}< 0\)

\(\frac{1}{7}-\frac{1}{6}< 0\)

\(...\)

\(\frac{1}{49}-\frac{1}{48}< 0\)

\(\frac{5}{6}\) khi cộng với các số nhỏ hơn 0 thì giá trị nó sẽ giảm, đồng thời còn bớt đi \(\frac{1}{50}\)

Do đó \(A< \frac{5}{6}\)

24 tháng 7 2016

\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}\)

\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)

\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)

\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\)

\(A=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}< 5.\frac{1}{25}+10.\frac{1}{30}+10.\frac{1}{40}\)

\(A< \frac{1}{5}+\frac{1}{3}+\frac{1}{4}< \frac{1}{4}+\frac{1}{3}+\frac{1}{4}=\frac{5}{6}\left(đpcm\right)\)

23 tháng 6 2019

\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}=\left(1+\frac{1}{3}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right).\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\right)\)\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}-\left(1+\frac{1}{2}+...+\frac{1}{25}\right)=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\)

\(A=\left(\frac{1}{26}+\frac{1}{27}+...+\frac{1}{35}\right)+\left(\frac{1}{36}+...+\frac{1}{50}\right)>\frac{1}{35}.10+\frac{1}{50}.15=\frac{41}{70}>\frac{7}{12}\)

\(A< \frac{10}{26}+\frac{15}{36}< \frac{5}{6}\) Vậy ....