B=( \(\frac{5}{2\cdot4}\)+ \(\frac{5}{4\cdot6}\)+ ....... +\(\frac{5}{100\cdot102}\))
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VA
25 tháng 7 2016
Đặt \(A=\frac{5}{2.4}+\frac{5}{4.6}+\frac{5}{6.8}+..+\frac{5}{100.102}\)
\(\frac{2}{5}A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{3}{6.8}+...+\frac{2}{100.102}\)
\(\frac{2}{5}A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{102}\)
\(\frac{2}{5}A=\frac{1}{2}-\frac{1}{102}\)
\(A=\frac{25}{51}:\frac{2}{5}\)
\(A=\frac{125}{102}\)
Ủng hộ mk nha !!! *_*
S
26 tháng 7 2016
\(\text{Đ}\text{ặt}:A=\frac{5}{2.4}+\frac{5}{4.6}+\frac{5}{6.8}+..+\frac{5}{100.102}\)
\(\frac{2}{5}A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{3}{6.8}+...+\frac{2}{100.102}\)
\(\frac{2}{5}A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{102}\)
\(\frac{2}{5}A=\frac{1}{2}-\frac{1}{102}\)
\(A=\frac{25}{51}:\frac{2}{5}\)
\(A=\frac{125}{102}\)
LN
3
11 tháng 2 2017
\(=\frac{\left(2.3.4.5\right)^2}{2.3^2.4^2.5.6}=\frac{2.5}{6}=\frac{5}{3}\)
11 tháng 2 2017
\(\frac{2^2}{1.3}=\frac{4}{3};\frac{3^2}{2.4}=\frac{9}{8};\frac{4^2}{3.5}=\frac{16}{15};\frac{5^2}{4.6}=\frac{25}{24}\)
\(\Rightarrow\frac{4}{3}.\frac{9}{8}.\frac{16}{15}.\frac{25}{24}=\frac{5}{3}\)
DP
24 tháng 6 2017
2, \(\frac{10}{1.2.3}+\frac{10}{2.3.4}+\frac{10}{3.4.5}+....+\frac{10}{100.101.102}\)
\(=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{102-100}{100.101.102}\)
\(=\frac{10}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{100.101}-\frac{1}{101.102}\right)\)
\(=\frac{10}{2}.\left(\frac{1}{1.2}-\frac{1}{101.102}\right)\)
\(=\frac{10}{2}.\frac{2575}{5151}\)
\(=2,499514657\)
30 tháng 4 2018
\(A=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}+\frac{2}{6.7}\)
\(A=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(A=2.\left(1-\frac{1}{7}\right)\)
\(A=2.\frac{6}{7}\)
\(A=\frac{12}{7}\)
AK
30 tháng 4 2018
\(A=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}+\frac{2}{6.7}\)
\(A=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(A=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{6}-\frac{1}{7}\right)\)
\(A=2.\left(1-\frac{1}{7}\right)\)
\(A=2.\left(\frac{7}{7}-\frac{1}{7}\right)\)
\(A=2.\frac{6}{7}\)
\(A=\frac{12}{7}\)
Chúc bạn học tốt !!!
LC
2
7 tháng 5 2018
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{5}+\frac{1}{6}\)
=\(1-\frac{1}{6}\)
=\(\frac{5}{6}\)
\(B=\frac{5}{2.4}+\frac{5}{4.6}+...+\frac{5}{100.102}\)
\(B=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{4}\right)+\frac{5}{2}.\left(\frac{1}{4}-\frac{1}{6}\right)+...+\frac{5}{2}.\left(\frac{1}{100}-\frac{1}{102}\right)\)
\(B=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{100}-\frac{1}{102}\right)\)
\(B=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{102}\right)\)
\(B=\frac{5}{2}.\frac{25}{51}\)
\(B=\frac{125}{102}\)
đè là gì vậy cho mỗi phép tính để đấy à coi để người khác nghĩ đề à