\(N=\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\frac{2^{12}.3^{15}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}=\frac{2^{12}.3^{15}+2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}=\frac{2^{12}.3^{10}\left(3^5+5\right)}{2^{11}.3^{11}\left(2.3-1\right)}\)

cho mk sửa lại xíu:

\(\frac{2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^{11}.5}=\frac{2.6}{3.5}=\frac{12}{15}=\frac{4}{5}\)

mk sai xíu. hihi, học tốt nhé !

\(\frac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}=\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.\left(6-1\right)}=\frac{2.6}{3.5}=\frac{4}{5}\)

\(\frac{\left(2^2\right)^6.\left(3^2\right)^5+2^9.3^9.120}{\left(2^3\right)^4.3^{12}-2^{11}.3^{11}}\)=\(\frac{2^{12}.3^{10}+2^9.3^9.120}{2^{12}.3^{12}-2^{11}.3^{11}}\)=\(\frac{120}{3^2-2^3.3^3}\)=\(\frac{120}{9-8.27}\)=\(\frac{120}{-207}\)=\(\frac{-40}{69}\)

tk cho mk nhe

\(\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
\(=\dfrac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\dfrac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{11}.3^{11}\left(2.3-1\right)}\)
\(=\dfrac{2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^{11}.5}\)
\(=\dfrac{2^{12}.3^{10}.6}{2^{11}.3^{11}.5}=\dfrac{2.6}{3.5}=\dfrac{4}{5}\)

Giải giúp em với mn

A=(2^2)^6.(3^2)^5+2^9.3^9.2^3.5.3/2^12.3^10+2^11.3^11

=2^12.3^10+2^12.3^10.5/2^12.3^10+2^11.3.3^10

=2^12.3^10.(1+5)/2^11.3^10.(2+3)

=2^11.2.6/2^11.5

=12/5

xin lỗi mình quên them dấu - 

ket qua bang 29 nha ban

ket qua bang -4/3 nha ban