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\(A=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{13.16}+\frac{1}{16.19}\)
\(A=\frac{1}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{16}-\frac{1}{19}\right)\)
\(A=\frac{1}{3}\cdot\left(1-\frac{1}{19}\right)\)
\(A=\frac{1}{3}\cdot\frac{18}{19}=\frac{6}{19}\)
\(B=\frac{1}{32}+\frac{1}{96}+\frac{1}{192}+\frac{1}{320}+\frac{1}{480}\)
\(B=\frac{1}{4\cdot8}+\frac{1}{8\cdot12}+\frac{1}{12\cdot16}+\frac{1}{16\cdot20}+\frac{1}{20\cdot24}\)
\(B=\frac{1}{4}\cdot\left(\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{12}+\frac{1}{12}-\frac{1}{16}+\frac{1}{16}-\frac{1}{20}+\frac{1}{20}-\frac{1}{24}\right)\)
\(B=\frac{1}{4}\cdot\left(\frac{1}{4}-\frac{1}{24}\right)\)
\(B=\frac{1}{4}\cdot\frac{5}{24}=\frac{5}{96}\)
\(A=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{13.16}+\frac{1}{16.19}\)
\(A=\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{16}-\frac{1}{19}\right)\)
\(A=\frac{1}{3}\left(1-\frac{1}{19}\right)\)
\(A=\frac{1}{3}.\frac{18}{19}\)
\(A=\frac{6}{19}\)
\(B=\frac{1}{32}+\frac{1}{96}+\frac{1}{192}+\frac{1}{320}+\frac{1}{480}\)
\(B=\frac{1}{4.8}+\frac{1}{8.12}+\frac{1}{12.16}+\frac{1}{16.20}+\frac{1}{20.24}\)
\(B=\frac{1}{4}\left(\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{12}+...+\frac{1}{20}-\frac{1}{24}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{4}-\frac{1}{24}\right)\)
\(B=\frac{1}{2}.\frac{5}{24}\)
\(B=\frac{5}{48}\)
B = \(\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{301.304}\)
B = \(\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{301}-\frac{1}{304}\right)\)
B = \(\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{304}\right)\)
B = \(\frac{1}{3}.\frac{75}{304}\)
B = \(\frac{25}{304}\)
\(B=\left(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{301.304}\right):3\)
\(\Rightarrow B=\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{301}-\frac{1}{304}\right):3\)
\(\Rightarrow B=\left(\frac{1}{4}-\frac{1}{304}\right):3\)
\(\Rightarrow B=\frac{75}{304}:3=\frac{25}{304}\)
\(A=\frac{1}{1.4}-\frac{1}{4.7}-\frac{1}{7.10}-...-\frac{1}{2011.2014}\)
\(A=\frac{1}{1.4}-\left(\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{2011.2014}\right)\)
Đặt \(B=\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{2011.2014}\)
\(B=\frac{1}{3}.\left(\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{2011.2014}\right)\)
\(B=\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{2011}-\frac{1}{2014}\right)\)
\(B=\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{2014}\right)\)
\(B=\frac{1}{3}.\frac{1005}{4028}=\frac{335}{4028}\)
\(A=\frac{1}{4}-\frac{335}{4028}=\frac{168}{1007}\)
A = \(\frac{1}{1.4}-\frac{1}{4.7}-\frac{1}{7.10}-...-\frac{1}{2011.2014}\)
A = 1 + \(\frac{1}{4}\) - \(\frac{1}{4}\) + \(\frac{1}{7}\) - \(\frac{1}{7}\) + \(\frac{1}{10}\) -....- \(\frac{1}{2011}\) + \(\frac{1}{2014}\)
A = 1 + \(\frac{1}{2014}\) = \(\frac{2015}{2014}\)
Sai đề : \(\frac{1}{2011.2014}\)
\(A=\frac{1}{1.4}-\frac{1}{4.7}-\frac{1}{7.10}-...-\frac{1}{2011.2014}\)
\(A=\frac{1}{1.4}-\left(\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{2011.2014}\right)\)
Đặt \(B=\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{2011.2014}\)
\(B=\frac{1}{3}\left(\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{2011.2014}\right)\)
\(B=\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{2011}-\frac{1}{2014}\right)\)
\(B=\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{2014}\right)\)
\(B=\frac{1}{3}.\frac{1005}{4028}=\frac{335}{4028}\)
\(A=\frac{1}{4}-\frac{335}{4028}=\frac{168}{1007}\)
Chúc bạn học tốt !!!
\(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{48.50}.\)
\(=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}....+\frac{2}{48.50}\right)\)
\(=\frac{1}{2}.\left(\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+...+\frac{50-48}{48.50}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+.....+\frac{1}{48}-\frac{1}{50}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{50}\right)\)
\(=\frac{1}{2}.\frac{12}{25}=\frac{6}{25}\)
\(B=\frac{3}{1.4}+\frac{3}{4.7}+....+\frac{3}{97.100}\)
\(=\frac{4-1}{1.4}+\frac{7-4}{4.7}+....+\frac{100-97}{97.100}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{97}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
\(C=\frac{8}{7.14}+\frac{8}{14.21}+....+\frac{8}{91.98}\)
\(=\frac{7}{8}.\left(\frac{7}{7.14}+\frac{7}{14.21}+...+\frac{7}{91.98}\right)\)
\(=\frac{7}{8}.\left(\frac{1}{7}-\frac{1}{14}+\frac{1}{14}-\frac{1}{21}+.....+\frac{1}{91}-\frac{1}{98}\right)\)
\(=\frac{7}{8}.\left(\frac{1}{7}-\frac{1}{98}\right)\)
\(=\frac{7}{8}.\frac{13}{98}=\frac{13}{112}\)
đây là toán lớp 5 cơ mà
a)A=\(\frac{1}{1x4}\)+\(\frac{1}{4x7}\)+...+\(\frac{1}{16x19}\)
A=\(\frac{1}{3}\)x3x(\(\frac{1}{1.4}\)+\(\frac{1}{4.7}\)+.......+\(\frac{1}{16.19}\)
A=\(\frac{1}{3}\)x(\(\frac{3}{1.4}\)+\(\frac{3}{4.7}\)+............+\(\frac{3}{16.19}\))
A=\(\frac{1}{3}\)x(1-1/4+1/4-1/7+......+1/13-1/16+1/16-1/19)
A=\(\frac{1}{3}\)x(1-\(\frac{1}{19}\))
A=\(\frac{1}{3}\)x\(\frac{18}{19}\)
A=\(\frac{6}{19}\)
câu b tương tự tách mẫu ra thôi