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\(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+\frac{2}{11\cdot13}+\frac{2}{13\cdot15}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)
\(=\frac{1}{3}-\frac{1}{15}\)
\(=\frac{4}{15}\)
Chúc bn hok giỏi !!!!!!!!! ^_^
\(\frac{2^2}{1x3}\)x \(\frac{4^2}{3x5}\)x \(\frac{6^2}{5x7}\) x \(\frac{8^2}{7x9}\)
= \(\frac{4}{3}\)x \(\frac{16}{15}\)x \(\frac{36}{35}\)x \(\frac{64}{63}\)
= \(1.486077098\)
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{2009.2011}\)
\(=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{2009.2011}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{2009}-\frac{1}{2011}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{2011}\right)=\frac{1}{2}.\frac{2008}{6033}=\frac{1004}{6033}\)
\(\frac{1}{3x5}+\frac{1}{5x7}+\frac{1}{7x9}+.....+\frac{1}{2009x2011}\)
\(=\frac{1.2}{3.5.2}+\frac{1.2}{5.7.2}+\frac{1.2}{7.9.2}+....+\frac{1.2}{2009.2011.2}\)
\(=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+.....+\frac{2}{2009.2011}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{2011}\right)\)
\(=\frac{1}{2}.\frac{2008}{6033}=\frac{2008}{12066}\)
\(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+........+\frac{2}{57.59}\)
\(B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-.........-\frac{1}{59}\)
\(B=1-\frac{1}{59}\)
\(B=\frac{59}{59}-\frac{1}{59}=\frac{58}{59}\)
Vậy B = \(\frac{58}{59}\)
Lưu ý: Dấu "." là dấu nhân
\(B=\frac{2}{1.3}+\frac{1}{3.5}+\frac{2}{5.7}+...+\frac{1}{57.59}\)
\(B=1.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{57}-\frac{5}{59}\right)\)
\(B=1.\left(1-\frac{1}{59}\right)\)
\(B=1.\frac{58}{59}\)
\(B=\frac{58}{59}\)
\(\frac{1}{1\times3}+\frac{2}{3\times5}+..........+\frac{2}{99\times101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+........+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{100}{101}\)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{101}\right)\)
\(=\frac{1}{2}.\frac{100}{101}\)
\(=\frac{50}{101}\)
\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{99\cdot101}\)
\(=2\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{99\cdot101}\right)\)
\(=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{99\cdot101}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)
\(=\frac{1}{1}-\frac{1}{101}=\frac{101}{101}-\frac{1}{101}=\frac{100}{101}\)
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
học trường gì
\(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{99\cdot101}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)
\(=\frac{1}{3}-\frac{1}{101}\)
\(=\frac{98}{303}\)
Tích mk nha bn !!!! ^_^