\(=\dfrac{-2^{12}\cdot3^{10}-2^{12}\cdot3^{10}\cdot5}{-2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}=\dfrac{2^{12}\cdot3^{10}\cdot6}{2^{11}\cdot3^{11}\cdot7}=\dfrac{2\cdot6}{3\cdot7}=\dfrac{12}{21}=\dfrac{4}{7}\)

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

Giải giúp em với mn

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

\(=\dfrac{2^{12}\cdot3^{10}+2^9\cdot2^3\cdot3^9\cdot3\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\)

\(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{11}\cdot3^{11}\cdot5}\)

\(=\dfrac{2^{12}\cdot3^{10}\cdot6}{2^{11}\cdot3^{11}\cdot5}=\dfrac{2\cdot6}{3\cdot5}=\dfrac{12}{15}=\dfrac{4}{5}\)

ta có : \(\dfrac{4^6.9^5+6^9.120}{-8^4.3^{12}-6^{11}}=\dfrac{4\left(4^5.9^5+5.6^{10}\right)}{-2^{12}.3^{12}-6^{11}}=\dfrac{4\left(2^{10}.3^{10}+5.6^{10}\right)}{-2^{12}.3^{12}-6^{11}}\)

\(=\dfrac{4\left(6^{10}+5.6^{10}\right)}{-6^{12}-6^{11}}=\dfrac{4.6^{11}}{-6^{11}\left(6+1\right)}=-\dfrac{4}{7}\)

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

\(=\dfrac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{-2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}\)

\(=-\dfrac{2^{12}\cdot3^{10}\left(1+5\right)}{2^{11}\cdot3^{11}\cdot\left(2\cdot3-1\right)}=\dfrac{-2}{3}\)

GIÚP MÌNH VỚI MN ƠI

\(C=\dfrac{6^3+3\cdot6^2+3^3}{13}=\dfrac{3^3\cdot8+3^3\cdot4+3^3}{13}=27\)

Giải:

\(10.\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)

\(=10.\dfrac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)

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

\(=10.\dfrac{2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^{11}\left(2.3-1\right)}\)

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

\(=10.\dfrac{2^{13}.3^{11}}{2^{11}.3^{11}.5}\)

\(=10.\dfrac{2^2}{5}\)

\(=2^3=8\)

Vậy ...

\(=\dfrac{5^3\left(2^3+2+1\right)}{55}-\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\)

\(=5^2-\dfrac{2^{12}\cdot3^{10}\cdot\left(1+5\right)}{2^{11}\cdot3^{11}\left(2\cdot3-1\right)}=25-\dfrac{2}{3}\cdot\dfrac{6}{5}\)

=25-4/5

=24,2

Bài lớp 7 à bạn khó quá

Mình giải rồi nhưng không ra

dễ

treen tử :phân tích ra thừa số nguyên tố rồi đặt thừa số chung vs số mũ nhỏ hơn

dưới mẫu : âm nhân âm = dương , phân tích ra thừa số nguyên tố

Sau đó rut gọn ra kết quả

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

\(=\dfrac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^9.2^3.3.5}{-\left(2^3\right)^4.3^{12}+\left(2.3\right)^{11}}\)

\(=\dfrac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{-2^{12}.3^{12}+2^{11}.3^{11}}\)

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

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

\(=\dfrac{2^{12}.3^{10}.6}{6^{11}-6^{12}}\)

\(=\dfrac{2^{12}.3^{10}.2.3}{6^{11}\left(1-6\right)}\)\(=\dfrac{2^{13}.3^{11}}{6^{11}.\left(-5\right)}\) \(=\dfrac{2^{11}.3^{11}.2^2}{6^{11}.\left(-5\right)}=\dfrac{6^{11}.4}{6^{11}.\left(-5\right)}=\dfrac{4}{-5}=-\dfrac{4}{5}\)

bài này mk làm ròi ak, tại do lỗi trang wed nên mới bị đăng lên

bài mk giải khác bạn ak