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

Giải giúp em với mn

\(A=\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\dfrac{\left(2^2\right)^6.\left(3^2\right)^5+6^9.6.2^2.5}{\left(2^3\right)^4.3^{12}-6^{11}}=\dfrac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}=\dfrac{2^{11}.3^{10}\left(2^1+2^1.5\right)}{2^{11}.3^{10}\left(2^1.3^2-1.3^1\right)}=\dfrac{2+10}{2.9-1.3}=\dfrac{12}{15}=\dfrac{4}{5}\)

Giúp mình với

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

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 !

\(N=\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)

\(N=\frac{4^6.9^5+6^9.120}{4^8.9^6-6^9.6^2}\)

\(N=\frac{36^5.4+36^4.6.6.20}{36^5.576-36^4.6.6.36}\)

\(N=\frac{36^5.4+36^5.20}{36^5.576-36^5.36}\)

\(N=\frac{36^5.\left(4+20\right)}{36^5.\left(576-36\right)}\)

\(N=\frac{36^5.24}{36^5.540}\)

\(N=\frac{24}{540}\)

\(N=\frac{2}{45}\)

Vậy \(N=\frac{2}{45}\)

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

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

tk cho mk nhe

Ta có:
        \(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)

\(=\frac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^9.120}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}\)

\(=\frac{2^{12}.3^{10}+2^9.3^9.120}{2^{12}.3^{12}-2^{11}.3^{11}}\)

\(=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)

\(=\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.\left(6-1\right)}\)

\(=\frac{2^{12}.3^{10}.6}{2^{11}.3^{11}.5}\)

\(=\frac{2.1.6}{1.3.5}\)

\(=\frac{12}{15}\)

\(=\frac{4}{5}\)

              







































Đúng 100% nhé !
 

........... là gì vậy

trrertre

=5^1+3+5+7+.....+49

Số các số hạng từ 1 đến 49 là (49-1):1+1=50

Tổng các số từ 1 đên 49 là      (49+1).50:2=1250

   =5^1250

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 - 

toi la Hai ha ha