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\(S=\frac{3}{2.3}+\frac{3}{3.6}+...+\frac{3}{2014.6039}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2013.2014}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2013}-\frac{1}{2014}\)
\(=1-\frac{1}{2014}=\frac{2013}{2014}\)
câu 2: \(S=\frac{25^{28^{ }}+25^{24}+...+25^2+25^2+1}{25^{28}.25^2+25^{24}.25^4+...+25^2+1}\)
rút gọn ta được
\(S=\frac{1}{25^4+1}\)
H=\(\frac{1\cdot2\cdot3+2\cdot4\cdot6+3\cdot6\cdot9+5\cdot10\cdot15}{1\cdot3\cdot6+2\cdot6\cdot12+3\cdot9\cdot18+5\cdot15\cdot30}=\frac{1.2.3+2^3.\left(1.2.3\right)+3^3.\left(1.2.3\right)+5^3.\left(1.2.3\right)}{1.3.6+2^3.\left(1.3.6\right)+3^3.\left(1.3.6\right)+5^3.\left(1.3.6\right)}=\frac{1.2.3.\left(1+2^3+3^3+5^3\right)}{1.3.6.\left(1+2^3+3^3+5^3\right)}=\frac{2}{6}=\frac{1}{3}\)
\(\frac{2^8\times6}{3^3\times5^4}\div\frac{8^3\times9}{5^3\times3^3}-\left(2^{14}+3^{19}\right)\left(3^{81}+5^{64}\right)\left(2^4-4^2\right)\)
\(=\frac{2^9\times3}{3^3\times5^4}\times\frac{5^3\times3^3}{2^9\times3^2}-\left(2^{14}+3^{19}\right)\left(3^{81}+5^{64}\right)\left(2^4-2^4\right)\)
\(=\frac{2^9\times3^4\times5^3}{3^5\times5^4\times2^9}-\left(2^{14}+3^{19}\right)\left(3^{81}+5^{64}\right)\times0\)
\(=\frac{1}{3\times5}-0\)
\(=\frac{1}{15}\)
d)\(\frac{2.3+4.6+14.21}{3.5+6.10+21.35}=\frac{2.3+2.2.6+2.7.21}{3.5+3.2.10+3.7.35}=\frac{2.3+2.12+2.147}{3.5+3.20+3.245}=\frac{2\left(3+12+147\right)}{3\left(5+20+245\right)}\)
\(=\frac{2.162}{3.270}=\frac{54}{135}=\frac{2}{5}\)
\(a.\frac{-2019.2018+1}{\left(-2017\right).\left(-2019\right)+2018}\)
\(=\frac{2019.\left(-2018\right)+1}{2019.2017+2018}\)
\(=\frac{2019.\left(-2018\right)+1}{2019.2018-1}\)
\(=-\frac{2018}{2018}\)
\(=-1\)
Ta có : \(S=\frac{3}{2\cdot3}+\frac{3}{3\cdot6}+\frac{3}{4\cdot9}+...+\frac{3}{6039\cdot2014}\)
\(S=3\cdot\left(\frac{3}{6\cdot3}+\frac{3}{9\cdot6}+\frac{3}{12\cdot9}+...+\frac{3}{6039\cdot6042}\right)\)
\(S=3\cdot\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{6039}-\frac{1}{6042}\right)\)
\(S=3\cdot\left(\frac{1}{3}-\frac{1}{6042}\right)\)
\(S=3\cdot\frac{671}{2014}\)
\(S=\frac{2013}{2014}\)