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Đặt A=\(\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+...+\dfrac{1}{98.100}\)
A=\(\left(\dfrac{1}{1.3}+...+\dfrac{1}{97.99}\right)+\left(\dfrac{1}{2.4}+...+\dfrac{1}{98.100}\right)\)
A=\(\left(\dfrac{1}{1}-\dfrac{1}{99}\right)+\left(\dfrac{1}{2}-\dfrac{1}{100}\right)\)
A=\(\dfrac{98}{99}-\dfrac{49}{100}\)
A=\(\dfrac{4949}{9900}\)
Mà \(\dfrac{3}{4}=\dfrac{7425}{9900}\)
Vậy A<\(\dfrac{3}{4}\)
a)\(\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{2008\cdot2010}\)
\(=2\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{2008\cdot2010}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)=2\cdot\dfrac{502}{1005}=\dfrac{1004}{1005}\)
b)\(\dfrac{\dfrac{3}{41}-\dfrac{12}{47}+\dfrac{27}{53}}{\dfrac{4}{41}-\dfrac{16}{47}+\dfrac{36}{53}}=\dfrac{3\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}{4\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}=\dfrac{3}{4}\)
a) gọi biểu thức đó là A
Ta có công thức \(\dfrac{a}{b.c}=\dfrac{a}{c-b}.\left(\dfrac{1}{b}-\dfrac{1}{c}\right)\)
Dựa vào công thức trên, ta có
\(A=\dfrac{4}{2}.\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+....+\dfrac{1}{2008}-\dfrac{1}{2009}\right)\)
\(A=\dfrac{4}{2}.\left(\dfrac{1}{2}-\dfrac{1}{2009}\right)\)
\(A=2.\left(\dfrac{2007}{4018}\right)=\dfrac{2007}{2009}\)
b) dễ quá bạn tự làm. (không phải mink không biết làm đâu nha)
\(\dfrac{5}{2.4}+\dfrac{5}{4.6}+\dfrac{5}{6.8}+...+\dfrac{5}{48.50}\)
= \(\dfrac{2}{2}.\left(\dfrac{5}{2.4}+\dfrac{5}{4.6}+\dfrac{5}{6.8}+....+\dfrac{5}{48.50}\right)\)
\(\)\(=\dfrac{5}{2}.\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+....+\dfrac{2}{48.50}\right)\)
\(=\dfrac{5}{2}.\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{48}-\dfrac{1}{50}\right)\)
=\(\dfrac{5}{2}.\left(\dfrac{1}{2}-\dfrac{1}{50}\right)\)
=\(\dfrac{5}{2}.\dfrac{12}{25}\)
=\(\dfrac{6}{5}\)=\(1\dfrac{1}{5}\)
Nếu bạn không biết cách giải bài này có thể bảo mình viết cách giải giúp!!!
Chúc bạn làm tốt!!!
\(\dfrac{5}{2.4}+\dfrac{5}{4.6}+\dfrac{5}{6.8}+...+\dfrac{5}{48.50}\)
=\(\dfrac{5}{2}.\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{48.50}\right)\)
=\(\dfrac{5}{2}.\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{48}-\dfrac{1}{50}\right)\)
=\(\dfrac{5}{2}.\left(\dfrac{1}{2}-\dfrac{1}{48}\right)\)
=\(\dfrac{5}{2}.\dfrac{23}{48}\) = \(\dfrac{115}{96}\)
Ta có :
\(F=\dfrac{4}{2.4}+\dfrac{4}{4.6}+..................+\dfrac{4}{2008.2010}\)
\(\Rightarrow F=2\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+.............+\dfrac{2}{2008.2010}\right)\)
\(\Rightarrow F=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+..............+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)
\(\Rightarrow F=2\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)\)
\(\Rightarrow F=2.\dfrac{502}{1005}=\dfrac{1004}{1005}\)
\(F=\dfrac{4}{2.4}+\dfrac{4}{4.6}+\dfrac{4}{6.8}+......+\dfrac{4}{2008.2010}\)
\(F=\dfrac{4}{2}\left(\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+.....+\dfrac{1}{2008.2010}\right)\)
\(F=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+.....+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)\(F=2\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)\)\(F=2.\dfrac{502}{1005}\)
\(F=\dfrac{1004}{1005}\)
A=\(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+...+\dfrac{1}{20\cdot22}\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{20}-\dfrac{1}{22}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{22}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{11}{22}-\dfrac{1}{22}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{5}{11}\)
\(=\dfrac{5}{22}\)
\(A=\dfrac{6}{2.4}+\dfrac{6}{4.6}+\dfrac{6}{6.8}+\dfrac{6}{8.10}+...+\dfrac{6}{30.32}+\dfrac{6}{32.34}\)
\(=6\left(\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+\dfrac{1}{8.10}+...+\dfrac{1}{30.32}+\dfrac{1}{32.34}\right)\)
\(=6\cdot\dfrac{2}{2}\left(\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+\dfrac{1}{8.10}+...+\dfrac{1}{30.32}+\dfrac{1}{32.34}\right)\)
\(=\dfrac{6}{2}\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+\dfrac{2}{8.10}+...+\dfrac{2}{30.32}+\dfrac{2}{32.34}\right)\)
\(=3\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{10}+...+\dfrac{1}{30}-\dfrac{1}{32}+\dfrac{1}{32}-\dfrac{1}{34}\right)\)
\(=3\left(\dfrac{1}{2}-\dfrac{1}{34}\right)=3\cdot\dfrac{8}{17}=\dfrac{24}{17}\)
A\(=6\left(\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+...+\dfrac{1}{32.34}\right)\)
A\(=6.\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{32}-\dfrac{1}{34}\right)\)
A\(=3\left(\dfrac{1}{2}-\dfrac{1}{34}\right)\)
A\(=3.\dfrac{8}{17}\)
A\(=\dfrac{24}{17}\)
\(\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{48.50}\)
\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{48.50}\right)\)
\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{4-2}{2.4}+\dfrac{6-4}{4.6}+...+\dfrac{50-48}{48.50}\right)\)
\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+..+\dfrac{1}{48}-\dfrac{1}{50}\right)\)
\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{50}\right)\)
\(\Leftrightarrow\dfrac{1}{2}.\dfrac{24}{50}=\dfrac{6}{25}\)
a: \(=\left(\dfrac{28}{42}+\dfrac{12}{42}-\dfrac{3}{42}\right):\left(\dfrac{-28}{28}-\dfrac{12}{28}+\dfrac{3}{28}\right)\)
\(=\dfrac{37}{42}:\dfrac{-37}{28}=\dfrac{-28}{42}=-\dfrac{2}{3}\)
b: \(=\dfrac{2+8+18+32+50}{12+48+108+192+300}=\dfrac{110}{660}=\dfrac{1}{6}\)
\(D=\dfrac{3}{2.4}+\dfrac{3}{4.6}+\dfrac{3}{6.8}+...+\dfrac{3}{98.100}\)
\(=\dfrac{3}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{98}-\dfrac{1}{100}\right)\)
\(=\dfrac{3}{2}\left(\dfrac{1}{2}-\dfrac{1}{100}\right)\)
\(=\dfrac{3}{2}.\dfrac{49}{100}=\dfrac{147}{200}\)
\(D=\dfrac{3}{2\cdot4}+\dfrac{3}{4\cdot6}+\dfrac{3}{6\cdot8}+...+\dfrac{3}{98\cdot100}\\ =\dfrac{3}{2}\cdot\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{98\cdot100}\right)\\ =\dfrac{3}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{98}-\dfrac{1}{100}\right)\\ =\dfrac{3}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{100}\right)\\ =\dfrac{3}{2}\cdot\dfrac{49}{100}\\ =\dfrac{147}{200}\)