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\(A=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{13.16}+\frac{1}{16.19}\)
\(A=\frac{1}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{16}-\frac{1}{19}\right)\)
\(A=\frac{1}{3}\cdot\left(1-\frac{1}{19}\right)\)
\(A=\frac{1}{3}\cdot\frac{18}{19}=\frac{6}{19}\)
\(B=\frac{1}{32}+\frac{1}{96}+\frac{1}{192}+\frac{1}{320}+\frac{1}{480}\)
\(B=\frac{1}{4\cdot8}+\frac{1}{8\cdot12}+\frac{1}{12\cdot16}+\frac{1}{16\cdot20}+\frac{1}{20\cdot24}\)
\(B=\frac{1}{4}\cdot\left(\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{12}+\frac{1}{12}-\frac{1}{16}+\frac{1}{16}-\frac{1}{20}+\frac{1}{20}-\frac{1}{24}\right)\)
\(B=\frac{1}{4}\cdot\left(\frac{1}{4}-\frac{1}{24}\right)\)
\(B=\frac{1}{4}\cdot\frac{5}{24}=\frac{5}{96}\)
\(A=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{13.16}+\frac{1}{16.19}\)
\(A=\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{16}-\frac{1}{19}\right)\)
\(A=\frac{1}{3}\left(1-\frac{1}{19}\right)\)
\(A=\frac{1}{3}.\frac{18}{19}\)
\(A=\frac{6}{19}\)
\(B=\frac{1}{32}+\frac{1}{96}+\frac{1}{192}+\frac{1}{320}+\frac{1}{480}\)
\(B=\frac{1}{4.8}+\frac{1}{8.12}+\frac{1}{12.16}+\frac{1}{16.20}+\frac{1}{20.24}\)
\(B=\frac{1}{4}\left(\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{12}+...+\frac{1}{20}-\frac{1}{24}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{4}-\frac{1}{24}\right)\)
\(B=\frac{1}{2}.\frac{5}{24}\)
\(B=\frac{5}{48}\)
\(A=\frac{1}{1.4}-\frac{1}{4.7}-\frac{1}{7.10}-...-\frac{1}{2011.2014}\)
\(A=\frac{1}{1.4}-\left(\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{2011.2014}\right)\)
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) A = \(\frac{5}{1.4}+\frac{29}{4.7}+\frac{71}{7.10}+....+\frac{10301}{100.103}\) (có 34 số hạng)
A = \(\frac{4+1}{1.4}+\frac{4.7+1}{4.7}+\frac{7.10+1}{7.10}+....+\frac{100.103+1}{103.100}\)
A = \(1+\frac{1}{1.4}+1+\frac{1}{4.7}+1+\frac{1}{7.10}+....+1+\frac{1}{100.103}\)
A = \(1.34+\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}\right)\)
A = \(34+\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)
A = \(34+\frac{1}{3}.\left(1-\frac{1}{103}\right)\)
A = \(34+\frac{1}{3}\cdot\frac{102}{103}\)
A = \(34+\frac{34}{103}=\frac{3536}{103}\)
a, \(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{x\left(x+1\right)}=\frac{13}{90}\)
⇒ \(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{13}{90}\)
⇒ \(\frac{1}{5}-\frac{1}{x+1}=\frac{13}{90}\)
⇒ \(\frac{1}{x+1}=\frac{1}{5}-\frac{13}{90}\)
⇒ \(\frac{1}{x+1}=\frac{18}{90}-\frac{13}{90}\)
⇒ \(\frac{1}{x+1}=\frac{1}{18}\)
⇒ x + 1 = 18
⇒ x = 17
Vậy x = 17
b, \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}=\frac{49}{148}\)
⇒ \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x\left(x+3\right)}=\frac{49.3}{148}\)
⇒ \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{147}{148}\)
⇒ \(1-\frac{1}{x+3}=\frac{147}{148}\)
⇒ \(\frac{1}{x+3}=1-\frac{147}{148}\)
⇒ \(\frac{1}{x+3}=\frac{1}{148}\)
⇒ x + 3 = 148
⇒ x = 145
Vậy x = 145
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=\(\frac{6}{19}\)
\(\frac{45}{188}\)
=\(\frac{31}{94}\)theo mình mới giải là vậy còn nếu không bạn có thể trình bày cách giải cho mình luôn được không?