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a: \(A=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{201}-\dfrac{1}{203}=\dfrac{1}{3}-\dfrac{1}{203}=\dfrac{200}{609}\)
b: \(B=\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{73}-\dfrac{1}{76}\)
\(=\dfrac{1}{4}-\dfrac{1}{76}=\dfrac{18}{76}=\dfrac{9}{38}\)
\(=\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{2021}-\dfrac{1}{2024}=\dfrac{1}{4}-\dfrac{1}{2024}=\dfrac{505}{2024}\)
\(\dfrac{3}{4.7}+\dfrac{3}{7.10}+\dfrac{3}{10.13}+...+\dfrac{1}{2021.2024}\)
=\(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{14}+...+\dfrac{1}{2021}-\)\(\dfrac{1}{2024}\)
=\(\dfrac{1}{4}-\dfrac{1}{2024}\)
=\(\dfrac{505}{2024}\)
\(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+\dfrac{3}{10\cdot13}+\dfrac{3}{13\cdot16}\)
\(=\dfrac{3\cdot1}{1\cdot4}+\dfrac{3\cdot1}{4\cdot7}+\dfrac{3\cdot1}{7\cdot10}+\dfrac{3\cdot1}{10\cdot13}+\dfrac{3\cdot3}{13\cdot16}\)
\(=3\cdot\left(\dfrac{1}{1\cdot4}+\dfrac{1}{4\cdot7}+\dfrac{1}{7\cdot10}+\dfrac{1}{10\cdot13}+\dfrac{1}{13\cdot16}\right)\)
\(=3\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}\right)\)
\(=3\cdot\left(1-\dfrac{1}{16}\right)\)
\(=3\cdot\left(\dfrac{16}{16}-\dfrac{1}{16}\right)\)
\(=3\cdot\dfrac{15}{16}\)
\(=\dfrac{45}{16}\)
3/1.4+3/4.7+3/7.10+3/10.13
=1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + 1/10 - 1/13
=1 - 1/13
=12/13
\(A=3.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+..+\dfrac{3}{97.100}\right)\)
\(A=3.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
\(A=3.\left(1-\dfrac{1}{100}\right)=3.\dfrac{99}{100}=\dfrac{297}{100}\)
3/1.4 + 3/4.7 + .. +3/13.16
= 1/1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + 1/10 - 1/13 + 1/13 - 1/16
= 1/1 - 1/16
= 15/16
Ta có: \(c=\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+\frac{1}{10\cdot13}+....+\frac{1}{37\cdot40}\)
\(\Leftrightarrow3c=3\left(\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+\frac{1}{10\cdot13}+...+\frac{1}{37\cdot40}\right)\)
\(\Leftrightarrow3c=\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+...+\frac{3}{37\cdot40}\)
Mà \(\frac{3}{4\cdot7}=\frac{1}{4}-\frac{1}{7}\)
\(\frac{3}{7\cdot10}=\frac{1}{7}-\frac{1}{10}\)
...
\(\Leftrightarrow3c=\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+...+\frac{3}{37\cdot40}\)
\(=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{37}-\frac{1}{40}\)
Ta thấy ngoại trừ hai phân số đầu tiên và cuối cùng thì tất cả các phân số còn lại đều có 1 phân số có cùng giá trị tuyệt đối nhưng ngược dấu đứng cạnh, mà tổng hai số ngược dấu bằng 0 nên ta nhóm các phân số ngược dấu thì được:
\(3c=\frac{1}{4}-\frac{1}{40}\Leftrightarrow c=\left(\frac{1}{4}-\frac{1}{40}\right)\cdot\frac{1}{3}\)
\(=\frac{9}{40}\cdot\frac{1}{3}=\frac{3}{40}=\frac{9}{120}< \frac{40}{120}\)
Mà \(\frac{40}{120}=\frac{1}{3}\Rightarrow c< \frac{1}{3}\)
\(\dfrac{x}{1.4}\)+\(\dfrac{x}{4.7}+\dfrac{x}{7.10}+\dfrac{x}{10.13}+\dfrac{x}{13.16}=\dfrac{5}{2}\) \(x.\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}\right)=\dfrac{5}{2}\) \(x.\left(\dfrac{1}{1}-\dfrac{1}{16}\right)=\dfrac{5}{2}\) \(x.\dfrac{15}{16}=\dfrac{5}{2}\) \(x=\dfrac{5}{2}:\dfrac{15}{16}\) \(x=\dfrac{80}{30}=\dfrac{8}{3}\) DAY LA BAI LAM CUA MK NHO TICK CHO MK NHA CAM ON BAN TRUOC
\(=>C=\frac{3}{4}-\frac{3}{7}+\frac{3}{7}-\frac{3}{10}+....+\frac{3}{73}-\frac{3}{76}\)
\(=>C=\frac{3}{4}-\frac{3}{76}\)
\(=>C=\frac{54}{76}\)
\(=>C=\frac{27}{38}\)
\(C=\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+....+\frac{3}{73.76}\)
=> Cộng vế với vế , ta được :
\(C=\frac{3}{3}\left(\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+....+\frac{1}{73.76}\right)\)
\(\Rightarrow C=1\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{73}-\frac{1}{76}\right)\)
\(\Rightarrow C=1\left(\frac{1}{4}-\frac{1}{76}\right)\)
\(\Rightarrow C=1\left(\frac{19}{76}-\frac{1}{76}\right)=1\frac{18}{76}\)