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<=> \(\left(\frac{1}{3\cdot5}+\frac{1}{5.7}+...+\frac{1}{13\cdot15}\right)+x=\frac{17}{15}\)
<=> \(\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-...-\frac{1}{15}\right)+x=\frac{17}{15}\)
<=>\(\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{15}\right)+x=\frac{17}{15}\)
<=> \(\frac{2}{15}+x=\frac{17}{15}\)
=> x = 1
(1/3.5+1/5.7+1/7.9+1/9.11+1/11.13+1/13.15)+x=17/15
[2.(1/3-1/5+1/5-1/7+...+1/13-1/15)]+x=17/15
[2.(1/3-1/15)]+x=17/15
(2.4/15)+x=17/15
6/15+x=17/15
x=17/15-6/15
x=11/15
\(=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\)
\(=\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{9}\right)+\left(\frac{1}{9}-\frac{1}{11}\right)+\left(\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{1}{3}-\frac{1}{13}=\frac{10}{39}\)
Các bạn nêu rõ cách làm từng bài giúp mình nhé! Thanks ^-^!
B=2/1.3 + 2/3.5 + 2/5.7 +...+ 2/299.301
B=1-1/3+1/3-1/5+1/5-1/7+...+1/299-1/301=1-1/301=300/301
\(Ta có: \frac{2}{3}=\frac{1}{1}-\frac{1}{3}\);
\(\frac{2}{15}=\frac{1}{3}-\frac{1}{5}\);
\(\frac{2}{35}=\frac{1}{5}-\frac{1}{7}\) ; ... ; \(\frac{2}{89999}=\frac{1}{299}-\frac{1}{301}\).
=> B= \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{299}-\frac{1}{301}\)
=> B=\(\frac{1}{1}-\frac{1}{301}\)
=> B=\(\frac{300}{301}\)
Ta có : \(C=\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}+......+\frac{2}{41.42}\)
\(C=2\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+.....+\frac{1}{41.42}\right)\)
\(C=2\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....+\frac{1}{41}-\frac{1}{42}\right)\)
\(C=2\left(\frac{1}{3}-\frac{1}{42}\right)\)
\(C=2.\frac{13}{42}=\frac{13}{21}\)
2, \(\frac{10}{1.2.3}+\frac{10}{2.3.4}+\frac{10}{3.4.5}+....+\frac{10}{100.101.102}\)
\(=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{102-100}{100.101.102}\)
\(=\frac{10}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{100.101}-\frac{1}{101.102}\right)\)
\(=\frac{10}{2}.\left(\frac{1}{1.2}-\frac{1}{101.102}\right)\)
\(=\frac{10}{2}.\frac{2575}{5151}\)
\(=2,499514657\)
\(\frac{1}{3}+\frac{2}{15}+\frac{5}{35}+\frac{4}{63}+\frac{5}{99}\)
\(=\frac{1155}{3465}+\frac{462}{3465}+\frac{495}{3465}+\frac{220}{3465}+\frac{175}{3465}\)
\(=\frac{2507}{3465}\)
Đặt \(A=\frac{3}{3}+\frac{3}{15}+\frac{3}{35}+\frac{3}{63}+...+\frac{3}{143}\)
\(A=\frac{3}{1\cdot3}+\frac{3}{3\cdot5}+\frac{3}{5\cdot7}+...+\frac{3}{11\cdot13}\)
\(A=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...-\frac{1}{13}\right)\)
\(A=\frac{3}{2}\left(1-\frac{1}{13}\right)\)
\(A=\frac{3}{2}\cdot\frac{12}{13}=\frac{18}{13}\)
Vậy ...........
\(\frac{3}{3}+\frac{3}{15}+\frac{3}{35}+\frac{3}{63}+...+\frac{3}{143}.\)
\(=\frac{3}{1\times3}+\frac{3}{3\times5}+\frac{3}{5\times7}+\frac{3}{7\times9}+...+\frac{3}{11\times13}\)
\(=\frac{2}{3}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{11\times13}\right)\)
\(=\frac{2}{3}\times\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{2}{3}\times\left(1-\frac{1}{13}\right)\)
\(=\frac{2}{3}\times\frac{12}{13}\)
\(=\frac{8}{13}\)
\(C=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}\)
\(C=\frac{2}{3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\)
\(C=\frac{2}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\)
\(C=\frac{2}{3}+\frac{1}{3}-\frac{1}{13}\)
\(C=1-\frac{1}{13}\)
\(C=\frac{12}{13}\)
\(C=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\)
\(C=\frac{1}{1}-\frac{1}{3}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\)
\(C=\frac{1}{1}-\frac{1}{13}\)
\(C=\frac{12}{13}\)