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T
2 tháng 9 2020
\(\frac{5}{12}+\frac{5}{20}+\frac{5}{30}+...+\frac{5}{9900}=\frac{5}{3.4}+\frac{5}{4.5}+\frac{5}{5.6}+...+\frac{5}{99.100}\)
\(5\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(5\left(\frac{1}{3}-\frac{1}{100}\right)=\frac{97}{60}\)
Y
11 tháng 11 2017
Ta thấy đc quy luật:
\(\frac{2^2-1^2}{2^2}=\frac{2+1}{2+2}=\frac{3}{4}\)
\(\frac{2^2-1^2}{2^2}+\frac{3^2-2^2}{6^2}=\frac{6+2}{6+3}=\frac{8}{9}\)
\(\frac{2^2-1^2}{2^2}+\frac{3^2-2^2}{6^2}+\frac{4^2-3^2}{12^2}=\frac{12+3}{12+4}=\frac{15}{16}\)
Nên:
\(\frac{2^2-1^2}{2^2}+\frac{3^2-2^2}{6^2}+\frac{4^2-3^2}{12^2}+...+\frac{100^2-99^2}{9900^2}=\frac{9900+99}{9900+100}=\frac{9999}{10000}\)
Hay A<1(đpcm)
25 tháng 6 2016
(1 - 2/6) × (1 - 2/12) × (1 - 2/20) × ... × (1 - 2/9900)
= 4/6 × 10/12 × 18/20 × ... × 9898/9900
= 1.4/2.3 × 2.5/3.4 × 3.6/4.5 × ... × 98.101/99.100
= 1.2.3...98/3.4.5...100 × 4.5.6...101/2.3.4...99
= 2/99.100 × 100.101/2.3
= 101/99×3 = 101/297
☆☆☆☆☆
31 tháng 8 2020
a) \(\left(\frac{5}{25}-1,008\right):\frac{4}{7}:\left[\left(3\frac{1}{4}-6\frac{5}{9}\right)\cdot2\frac{2}{17}\right]\)
\(=\left(\frac{1}{5}-\frac{126}{125}\right):\frac{4}{7}:\left[\left(\frac{13}{4}-\frac{59}{9}\right)\cdot\frac{36}{17}\right]\)
\(=\left(\frac{25}{125}-\frac{126}{125}\right):\frac{4}{7}:\left[-\frac{119}{36}\cdot\frac{36}{17}\right]\)
\(=-\frac{101}{125}:\frac{4}{7}:\left(-7\right)=-\frac{101}{125}\cdot\frac{7}{4}\cdot\left(-\frac{1}{7}\right)=\frac{101}{500}\)
b) \(\left(-0,5-\frac{3}{5}\right):\left(-3\right)+\frac{1}{3}-\left(-\frac{1}{6}\right):\left(-2\right)\)
\(=\left(-\frac{1}{2}-\frac{3}{5}\right):\left(-3\right)+\frac{1}{3}-\left(-\frac{1}{6}\right)\cdot\left(-\frac{1}{2}\right)\)
\(=-\frac{11}{10}:\left(-3\right)+\frac{1}{3}-\frac{1}{12}\)
\(=\frac{11}{30}+\frac{1}{3}-\frac{1}{12}=\frac{37}{60}\)
26 tháng 12 2017
Ta có: \(\left(\frac{2017}{2}+\frac{2017}{6}+\frac{2017}{12}+...+\frac{2017}{9900}\right)\div\frac{99}{100}\)
\(=2017\cdot\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}\right)\cdot\frac{100}{99}\)
\(=2017\cdot\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\right)\cdot\frac{100}{99}\)
\(=2017\cdot\left(1-\frac{1}{100}\right)\cdot\frac{100}{99}\)
\(=2017\cdot\frac{99}{100}\cdot\frac{100}{99}\)
\(=2017\)
29 tháng 12 2017
Ta có\(\left(\frac{2017}{2}+\frac{2017}{6}+\frac{2017}{12}+...+\frac{2017}{9900}\right):\frac{99}{100}\)
Đặt B=\(\frac{2017}{2}+\frac{2017}{6}+\frac{2017}{12}+...+\frac{2017}{9900}\)
Ta có B =\(2017.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}\right)=2017.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)=2017.\left(1-\frac{1}{100}\right)=2017.\frac{99}{100}\)
Thay B vào A ta có A=\(2017.\frac{99}{100}:\frac{99}{100}=2017\)
a=\(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{99.100}\)
a=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
a=\(1-\frac{1}{100}\)
a=\(\frac{99}{100}\)
A = \(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{9900}\)
A = \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
A = \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
A = \(\frac{1}{1}-\frac{1}{100}\)
A = \(\frac{99}{100}\)
Vậy A = \(\frac{99}{100}\)
Hok tốt