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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)
\(=\frac{2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(2^2.3\right)^{10}}\)
\(=\frac{2^{19}.3^9+3.5.2^{18}.3^8}{2^9.3^9.2^{10}+2^{20}.3^{10}}\)
\(=\frac{2^{19}.3^9.\left(1+5\right)}{2^{19}.3^9.\left(1+2.3\right)}=\frac{6}{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\frac{2^{19}.27^9+15.4^9.9^4}{6^9.2^{12}+12^{10}}=\frac{2^{19}.\left(3^3\right)^9+3.5.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{12}+\left(2^2.3\right)^{10}}\)
\(=\frac{2^{19}.3^{27}+3.5.2^{18}.3^8}{2^9.3^9.2^{12}+2^{20}.3^{10}}=\frac{2^{19}.3^{27}+3^9.2^{18}.5}{2^{21}.3^9+2^{20}.3^{10}}=\frac{2^{18}.3^9.\left(2.3^{18}+5\right)}{2^{20}.3^9.\left(2+3\right)}\)
\(=\frac{1.1.\left(2.3^{18}+5\right)}{2^2.1.5}=\frac{2.3^{18}+5}{20}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\frac{2^{19}.3^9+3^9.5.2^{18}}{2^9.3^9.2^{10}+2^{20}.3^{10}}=\frac{2^{18}.3^9\left(2+5\right)}{2^{19}.3^9\left(1+2.3\right)}=\frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)
= \(\frac{2^{19}.3^9+5.2^{18}.3^9}{6^9.2^{10}+2^{10}.6^{10}}\)
=\(\frac{\left(2^{18}.3^9\right)\left(2+5\right)}{\left(6^9.2^{10}\right)\left(1+6\right)}\)
=\(\frac{7\left(2^{18}.3^9\right)}{7\left(3^9.2^9.2^{10}\right)}\)
= \(\frac{7\left(2^{18}.3^9\right)}{7\left(3^9.2^{19}\right)}\)
= \(\frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(C=\frac{2^{19}.3^9+5.2^{18}.3^9}{2^9.3^9.2^{10}+3^{10}.2^{20}}\)
\(C=\frac{3^9.\left(2^{19}+5.2^{18}\right)}{3^9.\left(2^9.2^{10}+3.2^{20}\right)}\)
\(C=\frac{2^{19}+5.2^{18}}{2^9.2^{10}+3.2^{20}}\)suy ra \(C=\frac{2^{18}.\left(2+5\right)}{2^{18}.\left(2+3.2^2\right)}\)
\(C=\frac{7}{14}\)
Vậy \(C=\frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}=\frac{2^{19}.3^9+5.2^{18}.3^9}{2^{10}.6^9+2^{10}.6^{10}}=\frac{2^{18}.3^9.\left(2+5\right)}{2^{10}.6^9\left(1+6\right)}=\frac{2^{18}.3^9.7}{2^{10}.6^9.7}=2^8.\left(\frac{1}{2}\right)^9=2^8.\frac{9}{2^9}=\frac{1}{2}.9=\frac{9}{2}\)Vậy C=\(\frac{9}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}=\frac{2^{19}.\left(3^3\right)^3+5.3.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(2^2.3\right)^{10}}\)
\(=\frac{2^{19}.3^9+5.3.2^{18}.3^8}{2^9.3^9.2^{10}+2^{20}.3^{10}}=\frac{2^{19}.3^9+5.2^{18}.3^9}{2^{19}.3^9+2^{20}.3^{10}}=\frac{2^{18}.3^9\left(2+5\right)}{2^{19}.3^9\left(1+2.3\right)}=\frac{2^{18}.3^9}{2^{19}.3^9}=\frac{1}{2}\)
P/s: Sai gì bỏ qua =)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\dfrac{2^{19}.\left(3^3\right)^3-3.5.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(3.2^2\right)^{10}}=\dfrac{2^{19}.3^9-5.2^{18}.3^9}{2^{19}.3^9+2^{20}.3^{10}}=\dfrac{2^{18}.3^9\left(2-5\right)}{2^{19}.3^9\left(1+6\right)}=\dfrac{-3}{2.7}=-\dfrac{3}{14}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)
\(=\frac{2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(3.2^2\right)^{10}}\)
\(=\frac{2^{19}.3^9+5.2^{18}.3^9}{2^{19}.3^9+3^{10}.2^{20}}\)
\(=\frac{2^{18}.3^9\left(2+5\right)}{2^{19}.3^9\left(1+3.2\right)}\)
\(=\frac{7}{2.7}=\frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2^{19}.27+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)
\(=\frac{2^{19}.3^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(2^2.3\right)^{10}}\)
\(=\frac{2^{19}.3^3+3.5.2^{18}.3^8}{2^9.3^9.2^{10}+2^{20}.3^{10}}\)
\(=\frac{2^{19}.3^3.\left(1+3^6.5\right)}{2^{19}.3^3.\left(3^6+2.3.3^6\right)}\)
\(=\frac{1+3^6.5}{3^6+2.3.3^6}\)
\(=\frac{1+3^6.5}{3^6+2.3.3^6}=\frac{3646}{5103}\)