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a) \(a=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{37\cdot39}\)
\(a=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}\)
\(a=\frac{1}{3}-\frac{1}{39}\)
\(a=\frac{12}{39}\)
b) \(\left(\frac{17}{28}+18.29+\frac{19}{30}+\frac{30}{31}\right)\cdot\left(\frac{-5}{12}+\frac{1}{4}+\frac{1}{6}\right)\)
\(=\left(\frac{17}{28}+18.29+\frac{19}{30}+\frac{30}{31}\right)\cdot\left(\frac{-5}{12}+\frac{3}{12}+\frac{2}{12}\right)\)
\(=\left(\frac{17}{28}+18.29+\frac{19}{30}+\frac{30}{31}\right)\cdot\left(\frac{-2}{12}+\frac{2}{12}\right)\)
\(=\left(\frac{17}{28}+18.29+\frac{19}{30}+\frac{30}{31}\right)\cdot0\)
\(=0\)
a) \(A=\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}+...+\frac{2}{37x39}\)
\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{37}-\frac{1}{39}\)
\(A=\frac{1}{3}-\frac{1}{39}\)
\(A=\frac{4}{13}\)
b) \(\left(\frac{17}{28}+18.29+\frac{19}{30}+\frac{30}{31}\right)x\left(\frac{-5}{12}+\frac{1}{4}+\frac{1}{6}\right)\)
\(=\left(\frac{17}{28}+18.29+\frac{19}{30}+\frac{30}{31}\right)x0\)
\(=0\)
Giải
Ta có A= [1+1/3.5] + [1+1/5.7] + [1+1/7.9] + ... + [1+1/37.39]
=>A= (1+1+1+...+1) +(1/3.5 + 1/5.7 + 1/7.9 + ... + 1/37.39)
=> A = 18 + 1/2.(2/3.5+2/5.7+2/7.9+...+2/37.39)
=>A = 18 + 1/2.(1/3-1/5+1/5-1/7+1/7-1/9+...+1/37-1/39)
=> A= 18 + 1/2.(1/3-1/39)
=> A= 18 + 1/2 . 4/13
=>A= 18 + 2/13 = 236/13
==> 2A = 2/3x5 + 2/5x7 +....+2/61x63
2A = 1/3-1/5+1/5-1/7+....+1/61-1/63
2A = 1/3-1/63
2A=20/63
==> A=10/63
A = \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
A = \(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)
A = \(\frac{1}{2}.\left(1-\frac{1}{11}\right)=\frac{1}{2}.\frac{10}{11}\)
A = \(\frac{5}{11}\)
\(\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+\dfrac{1}{7\cdot9}+...+\dfrac{1}{997\cdot999}\)
= \(\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{997}-\dfrac{1}{999}\right)\)
= \(\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{999}\right)\)
= \(\dfrac{1}{2}\cdot\dfrac{332}{999}=\dfrac{166}{999}\)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{101}\right)\)
\(=\frac{1}{2}.\frac{100}{101}\)
\(=\frac{50}{101}\)
\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{99\cdot101}\)
\(=2\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{99\cdot101}\right)\)
\(=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{99\cdot101}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)
\(=\frac{1}{1}-\frac{1}{101}=\frac{101}{101}-\frac{1}{101}=\frac{100}{101}\)
sửa đề tí nhé: \(x=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{197.199}\)
\(x=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{197}-\frac{1}{199}\right)\)
\(x=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{199}\right)\)
\(x=\frac{1}{2}.\frac{196}{597}\)
\(x=\frac{98}{597}\)
\(A=\dfrac{1}{3.5}+\dfrac{1}{7.9}+...+\dfrac{1}{37.39}\\ =\dfrac{1}{2}\left(\dfrac{2}{3.5}+\dfrac{2}{7.9}+...+\dfrac{2}{37.39}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{37}-\dfrac{1}{39}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{39}\right)\\ =\dfrac{1}{2}.\dfrac{4}{13}\\ =\dfrac{2}{13}\)
A=13.5+17.9+...+137.39=12(23.5+27.9+...+237.39)=12(13−15+15−17+...+137−139)=12(13−139)=12.413=213