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Đặt A
Ta có công thức :
\(\frac{a}{b.c}=\frac{a}{c-b}.\left(\frac{1}{b}-\frac{1}{c}\right)\)
Dựa vào công thức, ta có
\(A=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+......+\frac{1}{48}-\frac{1}{50}\right)\)
\(A=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{50}\right)=\frac{5}{2}.\left(\frac{12}{25}\right)=\frac{6}{5}\)
Ai thấy đúng thì ủng hộ nha !!!
a, \(\frac{5}{2.4}+\frac{5}{4.6}+\frac{5}{6.8}+...+\frac{5}{48.50}\)
=\(\frac{5}{2}\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{48.50}\right)\)
=\(\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{48}-\frac{1}{50}\right)\)
=\(\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{50}\right)\)=\(\frac{5}{2}.\frac{12}{25}\)=\(\frac{6}{5}\)
A=\((\frac{1}{3.8}+\frac{1}{8.13}+...+\frac{1}{33.38})\)
A=\(\frac{1}{5}\left(\frac{5}{3.8}+\frac{5}{8.13}+...+\frac{5}{33.38}\right)\)
A=\(\frac{1}{5}\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+...+\frac{1}{33}-\frac{1}{38}\right)\)
A=\(\frac{1}{5}.\left(\frac{1}{3}-\frac{1}{38}\right)\)
A=\(\frac{1}{5}.\frac{35}{114}\)
A=\(\frac{7}{114}\)
B=\((\frac{1}{3.10}+\frac{1}{10.17}+...+\frac{1}{31.38})\)
B=\(\frac{1}{7}\left(\frac{7}{3.10}+\frac{7}{10.17}+...+\frac{7}{31.38}\right)\)
B=\(\frac{1}{7}\left(\frac{1}{3}-\frac{1}{10}+\frac{1}{10}-\frac{1}{17}+...+\frac{1}{31}-\frac{1}{38}\right)\)
B=\(\frac{1}{7}\left(\frac{1}{3}-\frac{1}{38}\right)\)
B=\(\frac{1}{7}.\frac{35}{114}\)
B=\(\frac{5}{114}\)
⇒ \(\frac{A}{B}\)=\(\frac{7}{114}:\frac{5}{114}=\frac{7}{114}.\frac{114}{5}=\frac{7}{5}\)
Vậy \(\frac{A}{B}=\frac{7}{5}\)
A = \(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+....+\frac{1}{33}-\frac{1}{38}\)
=\(\frac{1}{3}-\frac{1}{38}\)
=\(\frac{35}{114}\)
B =\(\frac{1}{3}-\frac{1}{10}+\frac{1}{10}-\frac{1}{17}+...+\frac{1}{31}-\frac{1}{38}\)
=\(\frac{1}{3}-\frac{1}{38}\)
=\(\frac{35}{114}\)
=>tỉ số \(\frac{A}{B}\)= \(\frac{35}{114}:\frac{35}{114}\)=1
sách 6,7,8 có 2 bài này nè. mk k bt ghi ps nên mk ko gửi đc sorry nha. Hhh
a)\(A=\frac{10^{2014}+2016}{10^{2015}+2016}=>10A=\frac{10^{2015}+20160}{10^{2015}+2016}=1+\frac{18144}{10^{2015}+2016}\left(1\right)\)
\(B=\frac{10^{2015}+2016}{10^{2016}+2016}=>10B=\frac{10^{2016}+20160}{10^{2016}+2016}=1+\frac{18144}{10^{2016}+2106}\left(2\right)\)
từ 1 zà 2
=> 10A>10B
=>A>B
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10}\)
= \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+...+\frac{1}{10}\)
= \(1-\frac{1}{10}\)
= \(\frac{9}{10}\)
Bạn học tốt nhea ♥
a)\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{100.103}\\ =\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\\ =\frac{1}{3}.\left(1-\frac{1}{103}\right)\\ =\frac{1}{3}.\frac{102}{103}\\ =\frac{34}{103}\)
Câu 1 :
=\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}.\)
=\(\frac{1}{3}-\frac{1}{7}=\frac{4}{21}.\)
Câu 2 :
=\(\frac{23.23+6}{23.\left(23+1\right)-17}\)
=\(\frac{23.23+6}{23.23+23-17}\)
=\(\frac{23.23+6}{23.23+6}\)
=1.
A = \(\frac{3^2}{1\cdot4}+\frac{3^2}{4\cdot7}+\frac{3^2}{7\cdot10}+\frac{3^2}{10\cdot13}+\frac{3^2}{13\cdot16}+...+\frac{3^2}{97\cdot100}\)
A : 3 = \(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+\frac{3}{13\cdot16}+...+\frac{3}{97\cdot100}\)
A : 3 = \(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+...+\frac{1}{97}-\frac{1}{100}\)
A : 3 = \(\frac{1}{1}-\frac{1}{100}\)
A : 3 = \(\frac{99}{100}\)
A = \(\frac{297}{100}\)
\(A=\dfrac{1}{2\cdot6}+\dfrac{1}{3\cdot8}+...+\dfrac{1}{2023\cdot4048}\)
\(=2\left(\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+...+\dfrac{1}{4046\cdot4048}\right)\)
\(=2\cdot\dfrac{1}{2}\left(\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{4046\cdot4048}\right)\)
\(=\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{4046\cdot4048}\)
\(=\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{4046}-\dfrac{1}{4048}\)
\(=\dfrac{1}{4}-\dfrac{1}{4048}=\dfrac{1012}{4048}-\dfrac{1}{4048}=\dfrac{1011}{4048}\)
A=\(\frac{1}{2\cdot6}+\frac{1}{3\cdot8}+\frac{1}{4\cdot10}+\ldots+\frac{1}{2023\cdot4048}\)
A=\(\frac12\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\cdots+\frac{1}{2023\cdot2024}\right)\)
A=\(\frac12\left(\frac11-\frac12+\frac12-\frac13+\frac13-\frac14+\cdots+\frac{1}{2023}-\frac{1}{2024}\right)\)
A=\(\frac12\left(1-\frac{1}{2024}\right)\)
A=\(\frac12\cdot\frac{2023}{2024}\)
A=\(\frac{2023}{1012}\)