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\(C=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{5050}\)

\(=2\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{10100}\right)\)

\(=2\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{100\cdot101}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{100}-\dfrac{1}{101}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{101}\right)\)

\(=2\cdot\dfrac{99}{202}=\dfrac{99}{101}\)

16 tháng 8 2023

a) \(\dfrac{13}{20}+\dfrac{3}{5}+x=\dfrac{5}{6}\)

\(\Rightarrow\dfrac{5}{4}+x=\dfrac{5}{6}\)

\(\Rightarrow x=\dfrac{5}{6}-\dfrac{5}{4}\)

\(\Rightarrow x=\dfrac{-5}{12}\)

b) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\dfrac{-1}{3}\)

\(\Rightarrow x+\dfrac{1}{3}=\dfrac{11}{15}\)

\(\Rightarrow x=\dfrac{11}{15}-\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{2}{5}\)

c)\(\dfrac{-5}{8}-x=\dfrac{-3}{20}-\dfrac{-1}{6}\)

\(\dfrac{-5}{8}-x=\dfrac{1}{60}\)

\(\Rightarrow x=\dfrac{-5}{8}-\dfrac{1}{60}\)

\(\Rightarrow x=\dfrac{-77}{120}\)

d) \(\dfrac{3}{5}-x=\dfrac{1}{4}+\dfrac{7}{10}\)

\(\Rightarrow\dfrac{3}{5}-x=\dfrac{19}{20}\)

\(\Rightarrow x=\dfrac{3}{5}-\dfrac{19}{20}\)

\(\Rightarrow x=\dfrac{-7}{20}\)

e) \(\dfrac{-3}{7}-x=\dfrac{4}{5}+\dfrac{-2}{3}\)

\(\Rightarrow\dfrac{-3}{7}-x=\dfrac{2}{15}\)

\(\Rightarrow x=\dfrac{-3}{7}-\dfrac{2}{15}\)

\(\Rightarrow x=\dfrac{-59}{105}\)

16 tháng 8 2023

g) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)

\(\Rightarrow\dfrac{-5}{6}-x=\dfrac{1}{4}\)

\(\Rightarrow x=\dfrac{-5}{6}-\dfrac{1}{4}\)

\(\Rightarrow x=\dfrac{-13}{12}\)

19 tháng 3 2022

i giúp em vớiiiiii

 

\(M=\dfrac{3}{1+2}+\dfrac{3}{1+2+3}+...+\dfrac{3}{1+2+...+2022}\)

\(=\dfrac{3}{\dfrac{2\left(2+1\right)}{2}}+\dfrac{3}{\dfrac{3\left(3+1\right)}{2}}+...+\dfrac{3}{\dfrac{2022\left(2022+1\right)}{2}}\)

\(=\dfrac{6}{2\left(2+1\right)}+\dfrac{6}{3\left(3+1\right)}+...+\dfrac{6}{2022\cdot2023}\)

\(=\dfrac{6}{2\cdot3}+\dfrac{6}{3\cdot4}+...+\dfrac{6}{2022\cdot2023}\)

\(=6\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2022\cdot2023}\right)\)

\(=6\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\right)\)

\(=6\cdot\left(\dfrac{1}{2}-\dfrac{1}{2023}\right)=6\cdot\dfrac{2021}{4046}=\dfrac{12126}{4046}< 3\)

mà \(3< \dfrac{10}{3}\)

nên \(M< \dfrac{10}{3}\)

mn ghi giúp em chi tiết bài giải nx ạ! 

8 tháng 3 2022

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15 tháng 9 2021

a) \(\dfrac{1}{4}-3\left(\dfrac{1}{12}+\dfrac{3}{8}\right)=\dfrac{1}{4}-\dfrac{1}{4}-\dfrac{9}{8}=-\dfrac{9}{8}\)

b) \(\left(-\dfrac{2}{3}+\dfrac{3}{5}\right):\dfrac{1}{50}-30=\left(-\dfrac{2}{3}+\dfrac{3}{5}\right).50-30=-\dfrac{100}{3}+30-30=-\dfrac{100}{3}\)

27 tháng 2 2022

a, \(\frac{3}{10}+\frac{5}{8}=\frac{24}{80}+\frac{50}{80}=\frac{74}{80}=\frac{37}{40}\)

b, \(\frac{1}{8}+\frac{5}{6}=\frac{3}{24}+\frac{20}{24}=\frac{23}{24}\)

c, \(\frac{1}{3}+\frac{1}{6}+\frac{1}{8}=\frac{8}{24}+\frac{4}{24}+\frac{3}{24}=\frac{15}{24}=\frac{5}{8}\)

#Fox

8 tháng 10 2021

đáp án là câu D:1/2

 

8 tháng 10 2021

D 1/2