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Từ công thức:\(1-\frac{2}{n.\left(n+1\right)}=\frac{n.\left(n+1\right)}{n.\left(n+1\right)}-\frac{2}{n.\left(n+1\right)}=\frac{n.\left(n+1\right)-2}{n.\left(n+1\right)}=\frac{n^2+n-2}{n.\left(n+1\right)}=\frac{n^2-n+2n-2}{n.\left(n+1\right)}\)
\(=\frac{n.\left(n-1\right)+2.\left(n-1\right)}{n.\left(n+1\right)}=\frac{\left(n+2\right).\left(n-1\right)}{n.\left(n+1\right)}\)
Ta có:B=\(\left(1-\frac{2}{42}\right)\left(1-\frac{2}{56}\right)...........\left(1-\frac{2}{2652}\right)\)
B=\(\left(1-\frac{2}{6.7}\right)\left(1-\frac{2}{7.8}\right).............\left(1-\frac{2}{51.52}\right)\)
B=\(\frac{5.8}{6.7}.\frac{6.9}{7.8}................\frac{50.53}{51.52}\)
B=\(\frac{5}{7}.\frac{53}{51}\)
B=\(\frac{265}{357}\)
\(A=\frac{12}{1.5}+\frac{12}{5.9}+\frac{12}{9.13}+.............+\frac{12}{101.105}\)
\(=3.\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+............+\frac{4}{101.105}\right)\)
\(=3\left(1-\frac{1}{105}\right)\)
\(=3.\frac{104}{105}=\frac{312}{105}\)
#)Trả lời :
\(A=\frac{\left(140+70+42+28+20+15\right)}{420}\)
\(A=\frac{315}{420}=\frac{\left(315:105\right)}{\left(420:105\right)}=\frac{3}{4}\)
Vậy : \(A=\frac{3}{4}\)
#~Will~be~Pens~#
đặt \(A=\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+........+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)
\(\frac{1}{2}A=\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+........+\frac{1}{x.\left(x+1\right)}=\frac{2}{9}.\frac{1}{2}\)
\(\frac{1}{2}A=\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+........+\frac{1}{x.\left(x+1\right)}=\frac{1}{9}\)
\(\frac{1}{2}A=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+.....+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)
\(\frac{1}{2}A=\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{18}\)
\(\Rightarrow x+1=18\)
\(\Rightarrow x=18-1\)
\(\Rightarrow x=17\)
vậy \(x=17\)
Sửa đề
\(R=\left(1-\frac{1}{21}\right)\left(1-\frac{1}{28}\right)\left(1-\frac{1}{36}\right)....\left(1-\frac{1}{1326}\right)\)
\(R=\frac{20}{21}.\frac{27}{28}.\frac{35}{36}....\frac{1325}{1326}\)
\(R=\frac{40}{42}.\frac{54}{56}.\frac{70}{72}.......\frac{2650}{2652}\)
\(R=\frac{5.8}{6.7}.\frac{6.9}{7.8}.\frac{7.10}{8.9}.......\frac{50.53}{51.52}\)
\(R=\frac{5.6.7.8.9...50}{6.7.8.9.10...51}.\frac{8.9.10...53}{7.8.9...52}=\frac{5}{51}.\frac{53}{7}=\frac{265}{357}\)
\(R=\frac{20}{21}.\frac{27}{28}.\frac{34}{35}....\frac{1325}{1326}\)
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