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\(a,=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(=\frac{1}{2}-0-0-0-...-0-\frac{1}{8}\)
\(=\frac{1}{2}-\frac{1}{8}\)
\(=\frac{4}{8}-\frac{1}{8}\)
\(=\frac{3}{8}\)
\(b,=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-...-\frac{1}{49}+\frac{1}{49}-\frac{1}{16}\)
\(=1-0-0-0-...-0-\frac{1}{16}\)
\(=1-\frac{1}{16}\)
\(=\frac{15}{16}\)
\(c,\frac{3}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-...-\frac{1}{51}\right)\)
\(=\frac{3}{2}.\left(1-0-0-0-...-\frac{1}{51}\right)\)
\(=\frac{3}{2}.\frac{50}{51}\)
\(=\frac{25}{17}\)
\(d,\)giống câu a tự làm nha mỏi tay quá.
\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}.\)
=> \(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{8}\)
=> \(A=\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)
\(B=\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{49.52}=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{49}-\frac{1}{52}\)
=> \(B=\frac{1}{4}-\frac{1}{52}=\frac{24}{104}=\frac{1}{26}\)
ta có A=3/1*3+3/3*5+3/5*7+...+3/49*51
=> A=3*1/2*(2/1*3+2/3*5+..+2/49*51)
=> A=3/2*(1-1/3+1/3-1/5+..+1/49-1/51)
=> A=3/2*(1-1/51)
=> A= 3/2* 50/51
=> A= 25/17
\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)
\(A=\frac{3}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\)
\(A=\frac{3}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(A=\frac{3}{2}.\left(1-\frac{1}{51}\right)\)
\(A=\frac{3}{2}.\frac{50}{51}=\frac{25}{17}\)
\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)
\(A=3.\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{49}-\frac{1}{50}\right)\)
\(A=\frac{3}{2}\left(1-\frac{1}{50}\right)\)
\(A=\frac{3}{2}.\frac{49}{50}\)
\(A=\frac{147}{100}\)
\(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{49.51}\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{51}\right)=\frac{1}{2}.\frac{50}{51}=\frac{25}{51}\)
5/2×(1/1-1/3+1/3-1/5+1/5-1/7+....+1/49-1/51)
5/2×(1/1-1/51)
5/2×50/51
2 và 23/51(hỗn số)
K cho mk nha bn, ok, cảm ơn bn nhìu
\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+....+\frac{3}{49.51}\)
\(=\frac{3}{2}\left(1-\frac{1}{3}+\frac{!}{3}-\frac{1}{5}+....+\frac{1}{49}-\frac{1}{51}\right)\)
\(=\frac{3}{2}\left(1-\frac{1}{51}\right)\)
\(=\frac{3}{2}.\frac{50}{51}=\frac{25}{17}\)
Đặt \(\)A = dãy trên
Ta có \(\frac{2}{3}A=\frac{2}{3}.\left(\frac{3}{1.3}+\frac{3}{3.5}+...+\frac{3}{49.51}\right)\)
\(=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{49.51}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{51}\)
\(=1-\frac{1}{51}\)
\(=\frac{50}{51}\)
\(\Rightarrow A=\frac{50}{51}\div\frac{2}{3}=\frac{25}{17}\)
Vậy kq của dãy là\(\frac{25}{17}\)
2/7A=2/1.3+2/3.5+...+2/99+101
2/7A=1-1/3+1/3-1/5+...+1/99-1/101
2/7A=1-1/101
2/7A=100/101
A=350/101
B=(1+1+1+1)-(1/2+1/6+1/12+1/20)
=4-(1/1.2+1/2.3+1/3.4+1/4.5)
=4-(1-1/2+1/2-1/3+1/3-1/4+1/4-1/5)
=4-(1-1/5)
=4-4/5
=16/5
a, \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\)
=2.(\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{99.101}\))
=\(2.\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
=\(\frac{2}{2}.\left(1-\frac{1}{101}\right)\)
\(=\frac{100}{101}\)
b, \(\frac{5}{1.3}+\frac{5}{3.5}+...+\frac{5}{99.101}\)
=\(5.\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{99.101}\right)\)
=\(5.\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\left(1-\frac{1}{101}\right)\)
=\(\frac{250}{101}\)
\(=\frac{5}{2}.\frac{100}{101}\)
a,21.321.3+23.523.5+25.725.7+....+299.101
=>\(\frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{101}\)
=>\(\frac{1}{1}-\frac{1}{101}\)
=>\(\frac{100}{101}\)
b,
51.351.3+53.553.5+55.755.7+....+599.101
=>\(\frac{5}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+....+\frac{2}{99.101}\right)\)
=>\(\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{101}\right)\)
=>\(\frac{5}{2}\left(\frac{1}{1}-\frac{1}{101}\right)\)
=>\(\frac{5}{2}.\frac{100}{101}\)
=>\(\frac{250}{101}\)
=3(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+..+1/49-1/51)
=3x50/51=50/17
theo tôi là thế còn các bn
M = \(3\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{49\cdot51}\right)\)\(=3\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{51}\right)\)
M = \(3\left(1-\frac{1}{51}\right)=3\cdot\frac{50}{51}=\frac{50}{17}\)