Tính B=\(\left(1-\dfrac{2}{2.3}\right)\).\(\left(1-\dfrac{2}{3.4}\right).\left(1-\dfrac{2}{4.5}\right)......\left(1-\dfrac{2}{99.100}\right)\)
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NN
25 tháng 7 2017
\(a,\dfrac{3}{4}-1\dfrac{1}{2}+0,5:\dfrac{5}{12}.\)
\(=\dfrac{3}{4}-\dfrac{3}{2}+\dfrac{1}{2}:\dfrac{5}{12}.\)
\(=\dfrac{3}{4}-\dfrac{6}{4}+\dfrac{1}{2}.\dfrac{12}{5}.\)
\(=-\dfrac{3}{4}+\dfrac{12}{10}.\)
\(=-\dfrac{3}{4}+\dfrac{6}{5}.\)
\(=-\dfrac{15}{20}+\dfrac{24}{20}=\dfrac{9}{20}.\)
Vậy.....
\(b,\left(-2\right)^2-1\dfrac{5}{27}.\left(-\dfrac{3}{2}\right)^3.\)
\(=4-1\dfrac{5}{27}.\left(-\dfrac{27}{8}\right).\)
\(=4-\dfrac{32}{27}.\left(-\dfrac{27}{8}\right).\)
\(=4-\left(-4\right).\)
\(=4+4=8.\)
Vậy.....
\(c,\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}.\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}.\)
\(=\dfrac{1}{2}+\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{99}-\dfrac{1}{99}\right)-\dfrac{1}{100}.\)
\(=\dfrac{1}{2}+0+0+...+0-\dfrac{1}{100}.\)
\(=\dfrac{1}{2}-\dfrac{1}{100}.\)
\(=\dfrac{50}{100}-\dfrac{1}{100}=\dfrac{49}{100}.\)
Vậy.....
HH
14 tháng 6 2018
Giải:
\(C=\left(1-\dfrac{2}{2.3}\right)\left(1-\dfrac{2}{3.4}\right)\left(1-\dfrac{2}{4.5}\right)...\left(1-\dfrac{2}{n\left(n+1\right)}\right)\)
Đk: \(n\ne0;n\ne-1\)
\(C=\left(1-\dfrac{2}{2.3}\right)\left(1-\dfrac{2}{3.4}\right)\left(1-\dfrac{2}{4.5}\right)...\left(1-\dfrac{2}{n\left(n+1\right)}\right)\)
\(\Leftrightarrow C=\left(\dfrac{2.3-2}{2.3}\right)\left(\dfrac{3.4-2}{3.4}\right)\left(\dfrac{4.5-2}{4.5}\right)...\left(\dfrac{n\left(n-1\right)-2}{n\left(n+1\right)}\right)\)
\(\Leftrightarrow C=\dfrac{4}{2.3}.\dfrac{10}{3.4}.\dfrac{18}{4.5}...\left(\dfrac{n\left(n-1\right)-2}{n\left(n+1\right)}\right)\)
\(\Leftrightarrow C=\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}.\dfrac{3.6}{4.5}...\left(\dfrac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\right)\)
\(\Leftrightarrow C=\dfrac{1.4.2.5.3.6...\left(n-1\right)\left(n+2\right)}{2.3.3.4.4.5.n\left(n+1\right)}\)
\(\Leftrightarrow C=\dfrac{\left[1.2.3...\left(n-1\right)\right]\left[4.5.6\left(n+2\right)\right]}{\left(2.3.4...n\right)\left[3.4.5....\left(n+1\right)\right]}\)
\(\Leftrightarrow C=\dfrac{n+2}{3n}\)
Vì \(\dfrac{n+2}{3n}< \dfrac{2n+2}{3n}\)
\(\Leftrightarrow C< \dfrac{2n+2}{3n}\)
Vậy ...
HH
14 tháng 6 2018
Giải:
\(C=\left(1-\dfrac{2}{2.3}\right)\left(1-\dfrac{2}{3.4}\right)\left(1-\dfrac{2}{4.5}\right)...\left(1-\dfrac{2}{n\left(n+1\right)}\right)\)
Đk: \(n\ne0;n\ne-1\)
\(C=\left(1-\dfrac{2}{2.3}\right)\left(1-\dfrac{2}{3.4}\right)\left(1-\dfrac{2}{4.5}\right)...\left(1-\dfrac{2}{n\left(n+1\right)}\right)\)
\(\Leftrightarrow C=\left(\dfrac{2.3-2}{2.3}\right)\left(\dfrac{3.4-2}{3.4}\right)\left(\dfrac{4.5-2}{4.5}\right)...\left(\dfrac{n\left(n-1\right)-2}{n\left(n+1\right)}\right)\)
\(\Leftrightarrow C=\dfrac{4}{2.3}.\dfrac{10}{3.4}.\dfrac{18}{4.5}...\left(\dfrac{n\left(n-1\right)-2}{n\left(n+1\right)}\right)\)
\(\Leftrightarrow C=\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}.\dfrac{3.6}{4.5}...\left(\dfrac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\right)\)
\(\Leftrightarrow C=\dfrac{1.4.2.5.3.6...\left(n-1\right)\left(n+2\right)}{2.3.3.4.4.5.n\left(n+1\right)}\)
\(\Leftrightarrow C=\dfrac{\left[1.2.3...\left(n-1\right)\right]\left[4.5.6\left(n+2\right)\right]}{\left(2.3.4...n\right)\left[3.4.5....\left(n+1\right)\right]}\)
\(\Leftrightarrow C=\dfrac{n+2}{3n}\)
Vì \(\dfrac{n+2}{3n}< \dfrac{2n+2}{3n}\)
\(\Leftrightarrow C< \dfrac{2n+2}{3n}\)
Vậy ...
18 tháng 9 2017
\(\left(1-\dfrac{2}{2.3}\right)\left(1-\dfrac{2}{3.4}\right)..............\left(1-\dfrac{2}{99.100}\right)\)
\(=\left(\dfrac{6}{2.3}-\dfrac{2}{2.3}\right).\left(\dfrac{12}{3.4}-\dfrac{2}{3.4}\right)..............\left(\dfrac{9900}{99.100}-\dfrac{2}{99.100}\right)\)
\(=\dfrac{4}{2.3}.\dfrac{10}{3.4}..........................\dfrac{9898}{99.100}\)
\(=\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}..............\dfrac{98.101}{99.100}\)
\(=\dfrac{1.2.3.....98}{2.3......99}.\dfrac{4.5.6.....101}{3.4.......100}\)
\(=\dfrac{1}{99}.\dfrac{101}{3}=\dfrac{101}{297}\)
MH
25 tháng 1 2022
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)
25 tháng 1 2022
b, \(\left(1-\dfrac{1}{100}\right)\left(1-\dfrac{1}{99}\right)...\left(1-\dfrac{1}{2}\right)=\dfrac{99.98...1}{100.99...2}=\dfrac{1}{100}\)
13 tháng 5 2022
11: \(=\left(1+\dfrac{1}{98}-1-\dfrac{1}{97}+\dfrac{1}{97}-\dfrac{1}{98}\right)\cdot\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}\right)=0\)
12: \(=\dfrac{7}{17}+\dfrac{10}{17}\cdot\left(\dfrac{-6+5}{10}\right)^2\)
\(=\dfrac{7}{17}+\dfrac{10}{17}\cdot\dfrac{1}{100}=\dfrac{7}{17}+\dfrac{1}{170}=\dfrac{71}{170}\)
14 tháng 6 2018
\(\Rightarrow\left(1+1+...+1\right)+2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...\dfrac{1}{n\left(n+1\right)}\right)\)[có (n-1) số 1]
\(\Rightarrow\left(n-1\right)+2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\right)\)
\(\Rightarrow\left(n-1\right)+2\left(\dfrac{1}{2}-\dfrac{1}{n+1}\right)\)
\(\Rightarrow\left(n-1\right)+\left(1-\dfrac{2}{n+1}\right)\)
\(\Rightarrow n-\dfrac{2}{n+1}\)
\(\Rightarrow\dfrac{n\left(n+1\right)}{n+1}-\dfrac{2}{n+1}\)
\(\Rightarrow\dfrac{n^2+n-2}{n+1}\)