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2 tháng 8 2016

\(=5.\left(\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{29.30}\right)\)

\(=5.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{29}-\frac{1}{30}\right)\)

\(=5.\left(\frac{1}{4}-\frac{1}{30}\right)\)

\(=5.\frac{13}{60}=\frac{13}{12}\)

2 tháng 8 2016

\(=5\left(\frac{1}{4\times5}+\frac{1}{5\times6}+...+\frac{1}{29\times30}\right)\)

\(=5\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{29}-\frac{1}{30}\right)\)

\(=5\left(\frac{1}{4}-\frac{1}{30}\right)\)

\(=5\times\frac{13}{60}\)

\(=\frac{13}{12}\)

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}.\left(\frac{1}{8.9}-\frac{1}{9.10}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)

\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right)=\frac{1}{2}.\frac{22}{45}=\frac{11}{45}\)

\(\frac{12}{35}:\frac{35}{25}=\frac{12}{35}.\frac{25}{35}=\frac{12.25}{35.35}=\frac{12.5.5}{7.5.7.5}=\frac{12}{49}\)

\(\frac{9}{22}.\frac{33}{18}=\frac{9.33}{22.18}=\frac{9.3.11}{11.2.9.2}=\frac{3}{4}\)

11 tháng 8 2017

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\) \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)

\(=1-\frac{1}{6}=\frac{5}{6}\)

11 tháng 8 2017

= 1 / 1   -   1 / 2   +   1 / 2   -   1 / 3   +   1 / 3   -   1 / 4   +   1 / 4   -   1 / 5   +   1 / 5   -   1 / 6

Ta gạch các ps trùng.

Còn lại :

1 / 1  -  1 / 6  =  6 / 5

21 tháng 3 2020

Ta có: 

\(\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+.....+\frac{1}{14\times15}=\frac{4-3}{3\times4}+\frac{5-4}{4\times5}+\frac{6-5}{6\times5}+...+\frac{15-14}{14\times15}\)

\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{14}-\frac{1}{15}=\frac{1}{3}-\frac{1}{15}=\frac{4}{15}\)

Do đó: \(\frac{4}{15}=\frac{x}{30}\)

               \(x=\frac{4}{15}\times30=8\)

21 tháng 3 2020

\(\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+...+\frac{1}{14\times15}=\frac{x}{30}\)

\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{5}+...+\frac{1}{14}-\frac{1}{15}=\frac{x}{30}\)

\(\frac{1}{3}-\frac{1}{15}=\frac{x}{30}\)

\(\frac{4}{15}=\frac{x}{30}\)

\(\Rightarrow4\times30=15\times x\)

\(120=15\times x\)

\(x=120\div15\)

\(x=8.\)

\(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{14.15}=\frac{x}{30}\)

\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{14}-\frac{1}{15}=\frac{x}{30}\)

\(\frac{1}{3}-\frac{1}{15}=\frac{x}{30}\)

\(\frac{5}{15}-\frac{1}{15}=\frac{x}{30}\)

\(\frac{4}{15}=\frac{x}{30}\)

\(\frac{8}{30}=\frac{x}{30}\)

\(x=8\)

hok tốt!!

4 tháng 7 2019

:V Làm sai hết rồi sai ngay từ bước đầu tiên.

\(\frac{1}{3.4}-\frac{1}{4.5}-\frac{1}{5.6}-....-\frac{1}{9.10}\)

\(=\frac{1}{3.4}-\left(\frac{1}{4.5}+\frac{1}{5.6}+....+\frac{1}{9.10}\right)\)

\(=\frac{1}{12}-\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{9}-\frac{1}{10}\right)\)

\(=\frac{1}{12}-\left(\frac{1}{4}-\frac{1}{10}\right)\)

\(=\frac{1}{12}-\frac{3}{20}\)

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

3 tháng 7 2019

\(\frac{1}{3.4}-\frac{1}{4.5}-...-\frac{1}{9.10}\)

\(-\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)

\(-\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(-\left(\frac{1}{3}-\frac{1}{10}\right)\)

\(-\frac{7}{30}\)