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\(1-\frac{1}{5.10}-\frac{1}{10.15}-\frac{1}{15.20}-...-\frac{1}{95.100}\)
\(=1-\left(\frac{1}{5.10}+\frac{1}{10.15}+\frac{1}{15.20}+...+\frac{1}{95.100}\right)\)
\(=1-\frac{1}{5}.\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{15}-\frac{1}{20}+...+\frac{1}{95}-\frac{1}{100}\right)\)
\(=1-\frac{1}{5}.\left(\frac{1}{5}-\frac{1}{100}\right)\)
\(=1-\frac{1}{5}.\frac{19}{100}\)
\(=1-\frac{19}{500}\)
\(=\frac{481}{500}\)
\(1-\frac{1}{5.10}-\frac{1}{10.15}-\frac{1}{15.20}-.....-\frac{1}{95.100}\)
\(=1-\left(\frac{1}{5.10}+\frac{1}{10.15}+\frac{1}{15.20}+...+\frac{1}{95.100}\right)\)
Đặt \(C=\frac{1}{5.10}+\frac{1}{10.15}+\frac{1}{15.20}+....+\frac{1}{95.100}\)
\(\Rightarrow C=\frac{1}{5}.\left(\frac{5}{5.10}+\frac{5}{10.15}+\frac{5}{15.20}+....+\frac{5}{95.100}\right)\)
\(=\frac{1}{5}.\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+....+\frac{1}{95}-\frac{1}{100}\right)\)
\(=\frac{1}{5}.\left(\frac{1}{5}-\frac{1}{100}\right)=\frac{1}{5}.\frac{19}{100}=\frac{19}{500}\)
\(\Rightarrow1-C=1-\frac{19}{500}=\frac{481}{500}\)
Chúc bạn học tốt
\(\frac{2}{5.10}+\frac{2}{10.15}+\frac{2}{15.20}+...+\frac{2}{2015.2020}\)
\(=2.\left(\frac{1}{5.10}+\frac{1}{10.15}+\frac{1}{15.20}+...+\frac{1}{2015.2020}\right)\)
\(=2.\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{15}-\frac{1}{20}+...+\frac{1}{2015}-\frac{1}{2020}\right)\)
\(=2.\left(\frac{1}{5}-\frac{1}{2020}\right)\)
\(=2.\frac{403}{2020}=\frac{403}{1010}\)
\(\frac{2}{5.10}+\frac{2}{10.15}+\frac{2}{15.20}+...+\frac{2}{2015.2020}\)
=\(\frac{2}{5}\left(\frac{5}{5.10}+\frac{5}{10.15}+\frac{5}{15.20}+...+\frac{5}{2015.2020}\right)\)
=\(\frac{2}{5}\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{15}-\frac{1}{20}+...+\frac{1}{2015}-\frac{1}{2016}\right)\)
=\(\frac{2}{5}.\left(\frac{1}{5}-\frac{1}{2020}\right)\)
=\(\frac{2}{5}.\frac{403}{2020}\)
=\(\frac{403}{5005}\)
mk chỉ làm thôi nhé
Ta tách biểu thức trên thành hai phần A và B
\(A=\frac{12-\frac{12}{7}-\frac{12}{289}-\frac{12}{85}}{4-\frac{4}{7}-\frac{4}{289}-\frac{12}{85}}\)
\(A=\frac{12.\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}{4.\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}\)
\(A=3\)
\(B=\frac{3+\frac{3}{13}+\frac{3}{169}+\frac{3}{91}}{7+\frac{7}{13}+\frac{7}{169}+\frac{7}{91}}\)
\(B=\frac{3.\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}{7.\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}\)
\(B=\frac{3}{7}\)
=> A:B=7
=>b=7
E = 2/5.10 + 2/10.15 + ... + 2/35.40
E = 2/5.(1/5 - 1/10 + 1/10 - 1/15 + ... + 1/35 - 1/40)
E = 2/5.(1/5 - 1/40)
E = 2/5.7/40
E = 7/100
E = \(\frac{2}{5.10}+\frac{2}{10.15}+...+\frac{2}{35.40}\)
= \(\frac{2}{5}.\left(\frac{5}{5.10}+\frac{5}{10.15}+...+\frac{5}{35.40}\right)\)
= \(\frac{2}{5}.\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{35}-\frac{1}{40}\right)\)
= \(\frac{2}{5}.\left(\frac{1}{5}-\frac{1}{40}\right)\)
= \(\frac{2}{5}.\frac{7}{40}\)
= \(\frac{7}{100}\)
\(\frac{1.2+3.6+5.10+7.14}{2.3+6.9+10.15+14.21}\)
\(=\frac{1.2+3.6+5.10+7.14}{1.2.3+3.6.3+5.10.3+7.14.3}\)
\(=\frac{1.2+3.6+5.10+7.14}{3.\left(1.2+3.6+5.10+7.14\right)}\)
\(=\frac{1}{3}\)
\(A=\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{29.32}+\frac{1}{32.35}\)
\(3A=\frac{3}{5.8}+\frac{3}{8.11}+....+\frac{3}{32.35}\)
\(3A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{32}-\frac{1}{35}\)
\(3A=\frac{1}{5}-\frac{1}{35}=\frac{6}{35}\)
\(A=\frac{6}{35}.\frac{1}{3}=\frac{2}{35}\)
\(B=\frac{1}{5.10}+\frac{1}{10.15}+...+\frac{1}{30.35}\)
\(5B=\frac{5}{5.10}+\frac{5}{10.15}+....+\frac{5}{30.35}\)
\(5B=\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{30}-\frac{1}{35}\)
\(5B=\frac{1}{5}-\frac{1}{35}=\frac{6}{35}\)
\(B=\frac{6}{35}.\frac{1}{5}=\frac{6}{175}\)
\(\frac{\left(\frac{6}{5}-\frac{4}{9}\right).\left(3^{22}+4.3^{18}\right)}{9^8.\left(x-2\right)}=\frac{\frac{2}{35}}{\frac{6}{175}}\)
\(\frac{\frac{34}{45}.3^{18}.\left(3^4+4\right)}{\left(3^2\right)^8.\left(x-2\right)}=\frac{5}{3}\)
\(\frac{\frac{34}{45}.85.3^{18}}{3^{16}.\left(x-2\right)}=\frac{5}{3}\)
\(\frac{\frac{578}{9}.3^2}{x-2}=\frac{5}{3}\)
\(\frac{578}{x-2}=\frac{5}{3}\)
\(\Rightarrow578.3=5x-10\)
\(\Rightarrow1734+10=5x\)
\(\Rightarrow x=\frac{1744}{5}=348,8\)
nếu có gì sai mấy bạn sửa nhé
n
\(a=3\left(\frac{1}{5.10}+\frac{1}{10.15}+...+\frac{1}{45.50}\right)\)
\(a=\frac{3}{5}\left(\frac{5}{5.10}+\frac{5}{10.15}+...+\frac{5}{45.50}\right)\)
\(a=\frac{3}{5}\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{45}-\frac{1}{50}\right)\)
\(a=\frac{3}{5}.\left(\frac{1}{5}-\frac{1}{50}\right)\)
\(a=\frac{3}{5}\cdot\frac{9}{50}\)
\(a=\frac{27}{250}\)