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Ta có : \(\frac{1}{10.9}-\frac{1}{9.8}-.....-\frac{1}{2.1}\)
\(=\frac{1}{90}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{9.8}\right)\)
\(=\frac{1}{90}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{8}-\frac{1}{9}\right)\)
\(=\frac{1}{90}-\left(1-\frac{1}{9}\right)\)
\(=\frac{1}{90}-\frac{8}{9}=\frac{-79}{90}\)
\(\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)
\(=\frac{2^{19}.\left(3^3\right)^3+15.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(2^2.3\right)^{10}}\)
\(=\frac{2^{19}.3^9+15.2^{18}.3^8}{2^9.3^9.2^{10}+2^{20}.3^{10}}\)
\(=\frac{2^{19}.3^9+15.2^{18}.3^8}{2^{19}.3^9+2^{20}.3^{10}}\)
\(=\frac{2^{18}.3^8\left(2.3+15\right)}{2^{19}.3^9\left(1+2.3\right)}\)
\(=\frac{6+15}{2.3\left(1+6\right)}\)
\(=\frac{21}{6.7}\)
\(=\frac{21}{42}\)
\(=\frac{1}{2}\)
Có P =\(\dfrac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}=\dfrac{2^{19}.\left(3^3\right)^3+5.3.\left(3^2\right)^4}{\left(2.3\right)^9+\left(3.2^2\right)^{10}}\)=\(\dfrac{2^{19}.3^9+5.3.2^{18}.3^8}{3^9.2^9.2^{10}+3^{10}.\left(2^2\right)^{10}}=\dfrac{2^{19}.3^9+5.2^{18}.3^9}{3^9.2^{19}+3^{10}.2^{20}}=\dfrac{2^{18}.3^9.\left(2+5\right)}{3^9.2^{19}.\left(1+3.2\right)}=\dfrac{2^{18}.3^9.7}{3^9.2^{19}.7}\)
=\(\dfrac{1}{2}\)
\(=\frac{2^{19}3^9+3\cdot5\cdot2^{18}\cdot3^8}{2^9\cdot3^9\cdot2^{10}+4^{10}\cdot3^{10}}=\frac{2^{19}\cdot3^9+5\cdot2^{18}\cdot3^9}{2^{19}\cdot3^9+2^{20}\cdot3^{10}}=\frac{2^{18}\cdot3^9\cdot\left(2+5\right)}{2^{19}\cdot3^9\left(1+6\right)}=\frac{1}{2}\)
\(\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}=\)\(\frac{5.2^{30}.3^{18}-4.3^{20}.2^{27}}{5.2^9.2^{19}.3^{19}-7.2^{29}.3^{18}}\)
\(=\frac{2^{27}.3^{18}.\left(5.2^3-4.3^2\right)}{2^{28}.3^{18}\left(5.1.3-7.2\right)}\)
\(=\frac{2^{27}.3^{18}.4}{2^{28}.3^{18}.1}\)
\(\frac{1.1.4}{2.1.1}=2\)