\(\frac{2^{12}.3^5-\left(2^2\right)^6.3^6}{2^{12}.\left(3^2\right)^3+\left(2^3\right)^4.3^3}\)

\(\frac{2^{12}.3^5.\left(1-3^{ }\right)}{2^{12}.3^3.\left(3^3-1\right)}\)

\(\frac{2^{12}.3^5.\left(-2\right)}{2^{12}.3^3.8}\)

\(\frac{3^2.\left(-1\right)}{4}\)

\(\frac{-9}{4}\)

VẬy.......................

nhớ tk cho mình nha

\(\dfrac{2^{12}.3^5-4^6.3^6}{2^{12}.9^3+8^4.3^5}=\dfrac{2^{12}.3^5-\left(2^2\right)^6.3^6.3}{2^{12}.\left(3^2\right)^3+\left(2^3\right)^4.3^5}.\)

\(=\dfrac{2^{12}.3^5-2^{12}.3^6}{2^{12}.3^6+2^{12}.3^5}=\dfrac{2^{12}\left(3^5-3^6\right)}{2^{12}\left(3^5+3^6\right)}=\dfrac{3^5-3^6}{3^5+3^6}=-\dfrac{486}{972}=-\dfrac{1}{2}.\)

Vậy..........

chịu

\(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}=\dfrac{2^{13}\cdot3^{11}}{2^{11}\cdot3^{11}\cdot7}=\dfrac{4}{7}\)

\(\dfrac{4^0\cdot9^3-6^9\cdot1200}{8^4\cdot3^{12}+6^{11}}\)

= \(\dfrac{1\cdot\left(3^2\right)^3-\left(3\cdot2\right)^9\cdot\left(5^2\cdot2^4\cdot3\right)}{\left(2^3\right)^4\cdot3^{12}+6^{11}}\)

= \(\dfrac{3^6-3^9\cdot2^9\cdot5^2\cdot2^4\cdot3}{2^{12}\cdot3^{12}+\left(2\cdot3\right)^{11}}\)

= \(\dfrac{3^6-3^{10}\cdot2^{13}\cdot5^2}{2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}\)

như lời đã nói thì bn làm tiếp>:)))