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\(N=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{9^2}\)
\(=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+....+\frac{1}{9.9}\)
\(N\)bé hơn \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{8.9}=N_1\)
\(N_1=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{8.9}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-.........-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)
\(=1-\frac{1}{9}\)
\(=\frac{8}{9}\) \((1)\)
\(N\)lớn hơn \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{9.10}=N_2\)
\(\Rightarrow N_2=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+......+\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-.....-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}\)
\(=\frac{5}{10}-\frac{1}{10}=\frac{2}{5}\) \((2)\)
Từ \((1)\)và \((2)\)suy ra ; \(\frac{2}{5}\)bé hơn N bé hơn \(\frac{8}{9}\)
Học tốt
Nhớ kết bạn với mình
\(=\frac{4}{5.7}+\frac{4}{7.9}+\frac{4}{9.11}+...+\frac{4}{59.61}\)
\(=2.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)\)
\(=2.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{59}-\frac{1}{61}\right)\)
=\(2.\left(\frac{1}{5}-\frac{1}{61}\right)\)
\(=2.\left(\frac{36}{505}\right)\)
\(=\frac{72}{505}\)
TK nha !!
Ta có : \(\frac{4}{5.7}+\frac{4}{7.9}+\frac{4}{9.11}+....+\frac{4}{59.61}\)
\(=2\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+.....+\frac{2}{59.61}\right)\)
\(=2\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+.....+\frac{1}{59}-\frac{1}{61}\right)\)
\(=2\left(\frac{1}{5}-\frac{1}{61}\right)\)
\(=2.\frac{56}{305}=\frac{112}{305}\)
\(4S=1+\frac{2}{4}+\frac{3}{4^2}+...+\frac{2019}{4^{2018}}\)
=> \(3S=1+\frac{2}{4}+\frac{3}{4^2}+...+\frac{2019}{2^{2018}}-\frac{1}{4}-\frac{2}{4^2}-\frac{3}{4^3}-...-\frac{2019}{4^{2019}}\)
=>3S=\(1+\frac{1}{4}+\frac{1}{4^2}+..+\frac{1}{2^{2018}}-\frac{2019}{4^{2019}}\)
còn lại tự giải nhé
\(\frac{1}{2!}+\frac{2!}{4!}+...+\frac{198!}{200!}=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{199.200}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-...+\frac{1}{199}-\frac{1}{200}=\left(\frac{1}{1}+\frac{1}{2}+...+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
1) \(\frac{3^{2014}.8^{19}}{6^{60}.3^{1955}}=\frac{3^{2014}.\left(2^3\right)^{19}}{\left(2.3\right)^{60}.3^{1955}}=\frac{3^{2014}.2^{57}}{2^{60}.3^{2015}}=\frac{1}{2^3.3}=\frac{1}{24}\)
2) \(5^x+5^{x+1}=150\)
=> 5x(1 + 5) = 150
=> 5x.6 = 150
=> 5x = 25
=> \(x=\pm2\)
3) \(\frac{3}{11.16}+\frac{3}{16.21}+...+\frac{3}{61.66}=\frac{3}{5}\left(\frac{5}{11.16}+\frac{5}{16.21}+...+\frac{5}{61.66}\right)\)
\(=\frac{3}{5}\left(\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+...+\frac{1}{61}-\frac{1}{66}\right)=\frac{3}{5}.\left(\frac{1}{11}-\frac{1}{66}\right)=\frac{3}{5}.\frac{5}{66}=\frac{1}{22}\)
\(\frac{4}{3.6}+\frac{4}{6.9}+...+\frac{4}{34.37}\)
\(=4\left(\frac{1}{3.6}+\frac{1}{6.9}+...+\frac{1}{34.37}\right)\)
\(=4.3\left(\frac{1}{3.6}+\frac{1}{6.9}+...+\frac{1}{34.37}\right):3\)
\(=4\left(\frac{3}{3.6}+\frac{3}{6.9} +...+\frac{3}{34.37}\right):3\)
\(=4.\frac{34}{111}:3\)
\(=\frac{136}{333}\)
\(\frac{4}{3.6}+\frac{4}{6.9}+...+\frac{4}{31.34}+\frac{4}{34.37}\)
\(=4\left(\frac{1}{3.6}+\frac{1}{6.9}+...+\frac{1}{31.34}+\frac{1}{34.37}\right)\)
\(=4\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{31}-\frac{1}{34}+\frac{1}{34}-\frac{1}{37}\right)\)
\(=4\left(\frac{1}{3}-\frac{1}{37}\right)\)
\(=4.\frac{34}{111}\)
\(=\frac{136}{111}\)