\(C=\frac{2^{19}.3^9+5.2^{18}.3^9}{2^9.3^9.2^{10}+3^{10}.2^{20}}\)

\(C=\frac{3^9.\left(2^{19}+5.2^{18}\right)}{3^9.\left(2^9.2^{10}+3.2^{20}\right)}\)

\(C=\frac{2^{19}+5.2^{18}}{2^9.2^{10}+3.2^{20}}\)suy ra \(C=\frac{2^{18}.\left(2+5\right)}{2^{18}.\left(2+3.2^2\right)}\)

\(C=\frac{7}{14}\)

Vậy \(C=\frac{1}{2}\)

treu ti thoi

problems?

\(\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}=\frac{2^{19}.\left(3^3\right)^3+5.3.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(2^2.3\right)^{10}}\)

\(=\frac{2^{19}.3^9+5.3.2^{18}.3^8}{2^9.3^9.2^{10}+2^{20}.3^{10}}=\frac{2^{19}.3^9+5.2^{18}.3^9}{2^{19}.3^9+2^{20}.3^{10}}=\frac{2^{18}.3^9\left(2+5\right)}{2^{19}.3^9\left(1+2.3\right)}=\frac{2^{18}.3^9}{2^{19}.3^9}=\frac{1}{2}\)

P/s: Sai gì bỏ qua =)

\(=\dfrac{2^{19}.\left(3^3\right)^3-3.5.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(3.2^2\right)^{10}}=\dfrac{2^{19}.3^9-5.2^{18}.3^9}{2^{19}.3^9+2^{20}.3^{10}}=\dfrac{2^{18}.3^9\left(2-5\right)}{2^{19}.3^9\left(1+6\right)}=\dfrac{-3}{2.7}=-\dfrac{3}{14}\)

\(=-\dfrac{3}{14}\)

\(\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)

\(=\frac{2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(2^2.3\right)^{10}}\)

\(=\frac{2^{19}.3^9+3.5.2^{18}.3^8}{2^9.3^9.2^{10}+2^{20}.3^{10}}\)

\(=\frac{2^{19}.3^9.\left(1+5\right)}{2^{19}.3^9.\left(1+2.3\right)}=\frac{6}{7}\)

st sai 1/2 mới đúng

\(=\frac{2^{19}.3^9+3^9.5.2^{18}}{2^9.3^9.2^{10}+2^{20}.3^{10}}=\frac{2^{18}.3^9\left(2+5\right)}{2^{19}.3^9\left(1+2.3\right)}=\frac{1}{2}\)

\(\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)

= \(\frac{2^{19}.3^9+5.2^{18}.3^9}{6^9.2^{10}+2^{10}.6^{10}}\)

=\(\frac{\left(2^{18}.3^9\right)\left(2+5\right)}{\left(6^9.2^{10}\right)\left(1+6\right)}\)

=\(\frac{7\left(2^{18}.3^9\right)}{7\left(3^9.2^9.2^{10}\right)}\)

= \(\frac{7\left(2^{18}.3^9\right)}{7\left(3^9.2^{19}\right)}\)

= \(\frac{1}{2}\)

\(\frac{2^{19}\times27^3+15\times4^9\times9^4}{6^9\times2^{10}+12^{10}}\)

\(=\frac{2^{19}\times\left(3^3\right)^3+5\times3\times\left(2^2\right)^9\times\left(3^2\right)^4}{\left(2\times3\right)^9\times2^{10}+\left(3\times4\right)^{10}}\)

\(=\frac{2^{19}\times3^9+3\times5\times2^{18}\times3^8}{3^9\times2^9\times2^{10}+3^{10}\times4^{10}}\)

\(=\frac{2^{19}\times3^9+5\times2^{18}\times3^9}{3^9\times2^{19}+3^{10}\times\left(2^2\right)^{10}}\)

\(=\frac{2^{18}\times3^9\times\left(2+5\right)}{3^9\times2^{19}+3^{10}\times2^{20}}\)

\(=\frac{2^{18}\times3^9\times7}{2^{19}\times3^9\times\left(1+3\times2\right)}\)

\(=\frac{7}{2\times\left(1+6\right)}\)

\(=\frac{7}{2\times7}\)

\(=\frac{1}{2}\)

 A = 2^19.27^3+15.4^9.9^4 / 6^9.2^10+12^10 
= 2^19.3^9 + 5.2^18.3^9 / 3^9.2^19 + 2^20.3^10 
= 2^18.3^9 ( 2 + 5 ) / 2^19.3^9.(1 + 2.3) 
= (2 + 5) / 2(1 + 6) 
= 7 / 2.7 
= 1/2

\(\frac{2^{19}.3^9+3^{13}.5.2^{18}}{2^{19}.3^9+2^{20}.3^{10}}=\frac{2^{18}\left(3^9+3^{13}.5\right)}{2^{19}\left(\cdot3^9+2.3^{10}\right)}=\frac{2^{18}.3^9\left(1+3^4.5\right)}{2^{19}.3^9\left(1+2.3\right)}=\frac{1+3^4.5}{2.7}=\frac{406}{14}=29\)

\(=\dfrac{2^{19}\cdot3^9+2^{18}\cdot3^9\cdot5}{2^{19}\cdot3^9+2^{20}\cdot3^{10}}=\dfrac{2^{18}\cdot3^9\cdot\left(2+5\right)}{2^{19}\cdot3^9\cdot7}=\dfrac{1}{2}\)