Bài 3 : 

\(A=\frac{1}{1\times2}+\frac{1}{2\times3}+....+\frac{1}{99\times100}\)

Ta có : \(\frac{1}{1\times2}=\frac{2-1}{1\times2}=\frac{2}{1\times2}-\frac{1}{1\times2}=1-\frac{1}{2}\)

           \(\frac{1}{2\times3}=\frac{3-2}{2\times3}=\frac{3}{2\times3}-\frac{2}{2\times3}=\frac{1}{2}-\frac{1}{3}\)

            \(\frac{1}{99\times100}=\frac{100-99}{99\times100}=\frac{100}{99\times100}-\frac{99}{99\times100}=\frac{1}{99}-\frac{1}{100}\)

  \(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=1-\frac{1}{100}\)

\(A=\frac{99}{100}\)

\(B=\frac{1}{10\times11}+\frac{1}{11\times12}+...+\frac{1}{38\times39}\)

Ta có : \(\frac{1}{10\times11}=\frac{11-10}{10\times11}=\frac{11}{10\times11}-\frac{10}{10\times11}=\frac{1}{10}-\frac{1}{11}\)

            \(\frac{1}{11\times12}=\frac{12-11}{11\times12}=\frac{12}{11\times12}-\frac{11}{11\times12}=\frac{1}{11}-\frac{1}{12}\)

           \(\frac{1}{38\times39}=\frac{39-38}{38\times39}=\frac{39}{38\times39}-\frac{38}{38\times39}=\frac{1}{38}-\frac{1}{39}\)

           \(\frac{1}{39\times40}=\frac{40-39}{39\times40}=\frac{40}{39\times40}-\frac{39}{39\times40}=\frac{1}{39}-\frac{1}{40}\)

\(\Rightarrow B=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+....+\frac{1}{38}-\frac{1}{39}+\frac{1}{39}-\frac{1}{40}\)

\(B=\frac{1}{10}-\frac{1}{40}\)

\(B=\frac{3}{40}\) 

3. 

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=1-\frac{1}{100}\)

\(A=\frac{99}{100}\)

\(B=\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{38.39}+\frac{1}{39.40}\)

\(B=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{38}-\frac{1}{39}+\frac{1}{39}-\frac{1}{40}\)

\(B=\frac{1}{10}-\frac{1}{40}\)

\(B=\frac{3}{40}\)

\(x\)là dấu nhân hả bạn? Nếu vậy thì mk làm cho nhé

\(A=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot....\cdot\left(1-\frac{1}{20}\right)\)

\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot.......\cdot\frac{17}{18}\cdot\frac{18}{19}\cdot\frac{19}{20}=\frac{1}{20}\)

Vậy \(A=\frac{1}{20}\)

\(B=1\frac{1}{2}\cdot1\frac{1}{3}\cdot1\frac{1}{4}\cdot........\cdot1\frac{1}{2005}\cdot1\frac{1}{2006}\cdot1\frac{1}{2007}\)

\(B=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot......\cdot\frac{2006}{2005}\cdot\frac{2007}{2006}\cdot\frac{2008}{2007}=\frac{2008}{2}=1004\)

Vậy \(B=1004\)

DẤU CHẤM LÀ DẤU NHÂN

a, 

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{19}{20}=\frac{1}{20}\)

b, \(1\frac{1}{2}.1\frac{1}{3}....1\frac{1}{2017}=\frac{3}{2}.\frac{4}{3}....\frac{2018}{2017}=\frac{2018}{2}=1009\)

???????????????????????????????????????????????????????????????????????????^_^

= 1/1 - 1/2 + 1/2 - 1/3 +.. + 1/5 - 1/6

= 1/1 - 1/6 

= 5/6

c=(1/2+1)x(1/3+1)x(1/4+1)x...x(1/99+1)

c=(1-1/2)x(1-1/3)x(1-1/4)x...x(1-1/99)

c1/2x2/3x3/4x...x89/99

c=1/99

\(A=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{99}{98}.\frac{100}{99}=\frac{3.4.5....99.100}{2.3.4...98.99}=\frac{100}{2}=50\)

=> A = 50

a) \(\frac{3}{5}+25-\frac{1}{5}=\left(\frac{3}{5}-\frac{1}{5}\right)+25=\frac{2}{5}+25=\frac{2}{5}+\frac{125}{5}=\frac{127}{5}\)

b) \(13\times3\times32,27+67,63\times39=39\times32,27+67,63\times39\)

                                                                 \(=39\times\left(32,27+67,63\right)\)

                                                                   \(=39\times99,9=3196,8\)

(Bạn xem lại đề nhé)

c) \(\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\times....\times\left(1-\frac{1}{100}\right)\)

\(=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times.....\times\frac{99}{100}\)

\(=\frac{1\times2\times3\times...\times99}{2\times3\times4\times....\times100}=\frac{1}{100}\)

a, \(\frac{3}{5}+25-\frac{1}{5}=\frac{127}{5}\)

b, \(\text{13 x 3 x 32,27 + 67,63 x 39 =}3896,1\)

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