LT
5
K
Khách
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SN
4
WH
6 tháng 5 2018
Trả lời
\(C=\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)
\(\Rightarrow C=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{...1}{49\cdot51}\)
\(\Rightarrow2C=2\left(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{49\cdot51}\right)\)
\(\Rightarrow2C=\frac{2}{1\cdot3}+\frac{2}{5\cdot7}+...+\frac{2}{49\cdot51}\)
\(\Rightarrow2C=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)
\(\Rightarrow2C=1-\frac{1}{51}\)
\(\Rightarrow2C=\frac{50}{51}\)
\(\Rightarrow C=\frac{50}{51}:2\)
\(\Rightarrow C=\frac{25}{51}\)
Vậy \(C=\frac{25}{51}\)
YN
30 tháng 3 2019
\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.\frac{2499}{2500}=\frac{3.8.15.2499}{4.9.16.2500}\)\(=\frac{14994}{24000}\)
(Thực hiện rút gọn)
# Học tốt #
30 tháng 3 2019
\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.\frac{2499}{2500}=\frac{3.8.15.2499}{4.9.16.2500}=\frac{3.15.2499}{4.9.2.2500}\)
Tự rút gọn tiếp đi
6 tháng 4 2019
=\(\frac{3}{2\cdot2}\cdot\frac{2\cdot4}{3\cdot3}\cdot\frac{3\cdot5}{4\cdot4}\cdot...\cdot\frac{49\cdot51}{50\cdot50}\)
=\(\frac{\left(2\cdot3\cdot...\cdot49\right)\cdot\left(3\cdot4\cdot...\cdot51\right)}{\left(2\cdot3\cdot4\cdot...\cdot50\right)\left(2\cdot3\cdot4\cdot...\cdot50\right)}\)
=\(\frac{51}{50\cdot2}=\frac{51}{100}\)
7 tháng 5 2019
\(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{49\cdot51}\)
\(\Rightarrow A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)
\(\Rightarrow A=\frac{1}{3}-\frac{1}{51}=\frac{17}{51}-\frac{1}{51}=\frac{16}{51}\)
\(B=5\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-...+\frac{1}{100}-\frac{1}{103}\right)\)
\(\Rightarrow B=5\cdot\left(1-\frac{1}{103}\right)=5\cdot\frac{102}{103}=\frac{510}{103}\)
\(C=5\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{101}\right)\)
\(\Rightarrow C=5\cdot\left(1-\frac{1}{101}\right)=5\cdot\frac{100}{101}=\frac{500}{101}\)
7 tháng 5 2019
\(B=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\)
\(B=\frac{5}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)
\(B=\frac{5}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(B=\frac{5}{3}\left(1-\frac{1}{103}\right)\)
\(B=\frac{5}{3}.\frac{102}{103}=\frac{170}{103}\)
\(C=\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}\)
\(C=\frac{5}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)
\(C=\frac{5}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(C=\frac{5}{2}\left(1-\frac{1}{101}\right)\)
\(C=\frac{5}{2}.\frac{100}{101}=\frac{250}{101}\)
KN
1 tháng 8 2020
\(M=\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)
\(\Rightarrow M=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)
\(\Rightarrow2M=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\)
\(\Rightarrow2M=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)
\(\Rightarrow2M=\frac{1}{3}-\frac{1}{51}\)
\(\Rightarrow2M=\frac{16}{51}\)
\(\Rightarrow M=\frac{8}{51}\)
\(N=\frac{-5}{1.3}+\frac{-5}{3.5}+...+\frac{-5}{2013.2015}\)
\(\Rightarrow N=-\frac{5}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2013.2015}\right)\)
\(\Rightarrow N=-\frac{5}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)
\(\Rightarrow N=-\frac{5}{2}\left(1-\frac{1}{2015}\right)\)
\(\Rightarrow N=-\frac{5}{2}.\frac{2014}{2015}\)
\(\Rightarrow N=-\frac{1007}{403}\)
DV
5 tháng 5 2017
\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{2}-\frac{1}{100}\)
\(A=\frac{49}{100}\)
DV
5 tháng 5 2017
\(B=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\)
\(B=\frac{5}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)
\(B=\frac{5}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(B=\frac{5}{3}.\left(\frac{1}{1}-\frac{1}{103}\right)\)
\(B=\frac{510}{103}\)
5 tháng 9 2020
\(a,\frac{3}{4}.\left(x+2\right)+\frac{1}{2}.\left(x-\frac{1}{2}\right)=\frac{15}{4}\)
\(\frac{3}{4}.x+\frac{3}{4}.2+\frac{1}{2}.x+\frac{1}{2}.\left(-\frac{1}{2}\right)=\frac{15}{4}\)
\(\left(\frac{3}{4}.x+\frac{1}{2}.x\right)+\frac{3}{2}-\frac{1}{4}=\frac{15}{4}\)
\(\left(\frac{3}{4}+\frac{1}{3}\right).x=\frac{15}{4}+\frac{1}{4}-\frac{3}{2}\)
\(\frac{5}{4}.x=\frac{5}{2}\)
\(x=\frac{5}{2}:\frac{5}{4}\)
\(x=2\)
\(b,3.x-\frac{3}{5}=0\)
\(3.x=0+\frac{3}{5}\)
\(3.x=\frac{3}{5}\)
\(x=\frac{3}{5}:3\)
\(x=\frac{1}{5}\)
\(c,\frac{-2}{3}.x-\frac{1}{3}.\left(2.x-3\right)=\frac{3}{2}\)
\(\frac{-2}{3}.x-\frac{2}{3}.x+1=\frac{3}{2}\)
\(\left(\frac{-2}{3}-\frac{2}{3}\right).x=\frac{3}{2}-1\)
\(-\frac{4}{3}.x=\frac{1}{2}\)
\(x=\frac{1}{2}:\left(\frac{-4}{3}\right)\)
\(x=\frac{-3}{8}\)
Học tốt
\(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}=\frac{1}{2}.(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51})\)
\(=\frac{1}{3}-\frac{1}{51}=\frac{16}{51}\)
\(C=\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)
\(C=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)
\(C=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)
\(C=\frac{1}{3}-\frac{1}{51}\)
\(C=\frac{16}{51}\)