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A=1/1-1/2+1/2-1/3+1/3-1/4+....+1/49-1/50
A=1/1-1/50
A=49/50
Vay A=49/50
B=1/3-1/5+1/5-1/7....+1/37-1/39
B=1/3-1/39
b=36/117
B=4/13
A = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
= \(1-\frac{1}{50}=\frac{49}{50}\)
B = \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}\)
= \(2\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{37.39}\right)\)
= \(2.\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}\right)\)
= \(\frac{2}{2}\left(\frac{1}{3}-\frac{1}{39}\right)\)
= \(\frac{4}{13}\)
C = \(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{73.76}\)
= \(3\left(\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{73.76}\right)\)
= \(3.\frac{1}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{73}-\frac{1}{76}\right)\)
= \(\frac{3}{3}\left(\frac{1}{4}-\frac{1}{76}\right)\)
= \(\frac{9}{38}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}\)
\(=\frac{49}{50}\)
\(S=\frac{5^2}{1.6}+\frac{5^2}{6.11}+...+\frac{5^2}{26.31}\)
\(S=5.\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{26.31}\right)\)
\(S=5.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{26}-\frac{1}{31}\right)\)
\(S=5.\left(1-\frac{1}{31}\right)\)
\(S=5.\frac{30}{31}\)
\(S=\frac{150}{31}\)
Câu L bạn thiếu số\(\frac{1}{475}\)
\(L=\frac{1}{7}+\frac{1}{91}+\frac{1}{247}+\frac{1}{475}+\frac{1}{775}+\frac{1}{1147}\)
\(L=\frac{1}{1.7}+\frac{1}{7.13}+\frac{1}{13.19}+\frac{1}{19.25}+\frac{1}{25.31}+\frac{1}{31.37}\)
\(L=\frac{1}{6}.\left(\frac{6}{1.7}+\frac{6}{7.13}+\frac{6}{13.19}+\frac{6}{19.25}+\frac{6}{25.31}+\frac{6}{31.37}\right)\)
\(L=\frac{1}{6}.\left(1-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{19}+\frac{1}{19}-\frac{1}{25}+\frac{1}{25}-\frac{1}{31}+\frac{1}{31}-\frac{1}{37}\right)\)
\(L=\frac{1}{6}.\left(1-\frac{1}{37}\right)\)
\(L=\frac{1}{6}.\frac{36}{37}\)
\(L=\frac{6}{37}\)
\(S=\frac{5^2}{1.6}+\frac{5^2}{6.11}+...+\frac{5^2}{26.31}\)
\(=5\left[\left(1-\frac{1}{6}\right)+\left(\frac{1}{6}-\frac{1}{11}\right)+...+\left(\frac{1}{26}-\frac{1}{31}\right)\right]\)
\(=5\left[1-\frac{1}{31}\right]\)
\(=5.\frac{30}{31}=\frac{150}{31}\)
\(L=\frac{1}{7}+\frac{1}{91}+\frac{1}{247}+\frac{1}{775}+\frac{1}{1147}\)
\(L.6=\frac{6}{1.7}+\frac{6}{7.13}+\frac{6}{13.19}+\frac{6}{19.25}+\frac{6}{25.31}\)
\(L.6=\left(1-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{13}\right)+\left(\frac{1}{13}-\frac{1}{19}\right)+\left(\frac{1}{19}-\frac{1}{25}\right)+\left(\frac{1}{25}-\frac{1}{31}\right)\)
\(L.6=1-\frac{1}{31}\)
\(L.6=\frac{31}{31}-\frac{1}{31}\)
\(L.6=\frac{25}{31}\)
\(L=\frac{30}{31}:6\)
\(L=\frac{30}{31}.\frac{1}{6}\)
\(L=\frac{30}{186}\)
a) \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{2017\cdot2018}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(=1-\frac{1}{2018}\)
\(=\frac{2017}{2018}\)
b) \(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{97\cdot99}\)( sửa 91.99 thành 97.99 mới đúng nha )
\(=\frac{1}{2}\left(\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+...+\frac{2}{97}-\frac{2}{99}\right)\)
\(=\frac{1}{2}\left(\frac{2}{3}-\frac{2}{99}\right)\)
\(=\frac{1}{2}.\frac{64}{99}\)
\(=\frac{32}{99}\)
a) 1/1.2 + 1/2.3 + 1/3.4 +...+1/2017.2018
= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ....+1/2017 - 1/2018
= 1 - 1/2018
= 2017/2018
\(A=\) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=1-\frac{1}{50}\)
\(A=\frac{49}{50}\)
\(A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.50}\)
A= \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)
A = \(\frac{1}{1}-\frac{1}{51}=\frac{50}{51}\)
x + 25 = 64
x = 64 - 25
x = 39
Vậy x = 39