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B = 3/4.8/9.15/16.....2499/2500
= 1.3/2^2 . 2.4/3^2 . 3.5/4^2 ..... 49.51/50^2
= 1.2.3.....49/1.2.3.....50 . 3.4.5....51/1.2.3.....50
= 1/50 . 51/2
= 51/100
Giải
B = 3/4.8/9.15/16.....2499/2500
B=1.3/2^2 . 2.4/3^2 . 3.5/4^2 ..... 49.51/50^2
B=1.2.3.....49/2.3.4.....50 . 3.4.5.....51/2.3.4.....50
B=1/50 . 51/2
B=51/100
\(A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{19.21}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\)
\(=\frac{1}{3}-\frac{1}{21}\)
\(=\frac{7}{21}-\frac{1}{21}=\frac{6}{21}\)
\(A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{19.21}\)
\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\)
\(A=\frac{1}{3}+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+\left(\frac{1}{9}-\frac{1}{9}\right)+...+\left(\frac{1}{19}-\frac{1}{19}\right)-\frac{1}{21}\)
\(A=\frac{1}{3}-\frac{1}{21}\)
\(A=\frac{2}{7}\)
a.
\(M=1.\left[\frac{1}{3}-\frac{1}{5}+.....\frac{1}{97}-\frac{1}{99}\right]\)
\(M=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
b.
\(N=\frac{3}{2}.\left[\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{197}-\frac{1}{199}\right]\)
\(N=\frac{3}{2}.\left[\frac{1}{5}-\frac{1}{199}\right]=\frac{291}{995}\)
mk đầu tiên nha bạn
G=\(\frac{3}{2.5}+\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{2015.2017}\)
G=\(3.\left(\frac{1}{2.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{2015.2017}\right)\)
G=\(3.\left(\frac{1}{2}.\frac{1}{5}+\frac{1}{5}.\frac{1}{7}+\frac{1}{7}.\frac{1}{9}+...+\frac{1}{2013}.\frac{1}{2015}+\frac{1}{2015}.\frac{1}{2017}\right)\)
G=\(3.\left(\frac{1}{2}+\frac{1}{2017}\right)\)
G=1.5
Anh ko bik có đúng ko nữa lâu quá rồi. Em thông cảm nhé
\(A=\dfrac{3}{3.5} + \dfrac{3}{5.7} + ... + \dfrac{3}{97.99}\)
\(\Rightarrow A=\dfrac{3}{2}\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{97.99}\right)\)
\(\Rightarrow A=\dfrac{3}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
\(\Rightarrow A=\dfrac{3}{2}\left(\dfrac{1}{3}-\dfrac{1}{99}\right)\)
\(\Rightarrow A=\dfrac{3}{2}.\dfrac{32}{99}\)
\(\Rightarrow A=\dfrac{16}{33}\)
Vậy \(A=\dfrac{16}{33}\)
A= \(\dfrac{3}{3.5}+\dfrac{3}{5.7}+\dfrac{3}{7.9}+...+\dfrac{3}{97.99}\)
= \(\dfrac{3}{2}.\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{97.99}\right)\)
= \(\dfrac{3}{2}.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
= \(\dfrac{3}{2}.\left(\dfrac{1}{3}-\dfrac{1}{99}\right)\)
= \(\dfrac{3}{2}.\dfrac{32}{99}\)
= \(\dfrac{3.32}{2.99}\)= \(\dfrac{3.2.3.6}{2.11.3.3}\)= \(\dfrac{6}{11}\)
B=\(\dfrac{3}{3.5}.\dfrac{3}{5.7}.....\dfrac{3}{47.49}\)
B=\(\dfrac{3}{2}.\left(\dfrac{2}{3.5}.\dfrac{2}{5.7}.....\dfrac{2}{47.49}\right)\)
B=\(\dfrac{3}{2}.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{47}-\dfrac{1}{49}\right)\)
B=\(\dfrac{3}{2}.\left(\dfrac{1}{3}-\dfrac{1}{49}\right)\)
B=\(\dfrac{3}{2}.\dfrac{46}{147}\)
B=\(\dfrac{23}{49}\)
a) Ta có: \(A=\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot...\cdot\dfrac{2499}{2500}\)
\(=\dfrac{1\cdot3}{2^2}\cdot\dfrac{2\cdot4}{3^2}\cdot\dfrac{3\cdot5}{4^2}\cdot...\cdot\dfrac{49\cdot51}{50^2}\)
\(=\dfrac{1}{50}\cdot\dfrac{51}{2}=\dfrac{51}{100}\)
b) Ta có: \(B=\dfrac{3}{3\cdot5}+\dfrac{3}{5\cdot7}+\dfrac{3}{7\cdot9}+...+\dfrac{3}{47\cdot49}\)
\(=\dfrac{3}{2}\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{47\cdot49}\right)\)
\(=\dfrac{3}{2}\left(\dfrac{1}{3}-\dfrac{1}{49}\right)\)
\(=\dfrac{3}{2}\cdot\dfrac{46}{147}=\dfrac{138}{294}=\dfrac{23}{49}\)