\(\dfrac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}=\dfrac{\left(2^4\right)^3.3^{10}+3.5.2^3.\left(2.3\right)^9}{\left(2^2\right)^6.3^{12}+\left(2.3\right)^{11}}\)

\(=\dfrac{2^{12}.3^{10}+3.5.2^3.2^9.3^9}{2^{12}.3^{12}+2^{11}.3^{11}}=\dfrac{2^{12}.3^{10}+5.2^{12}.3^{10}}{2^{12}.3^{12}+2^{11}.3^{11}}\)

\(=\dfrac{2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^{11}.\left(2.3+1\right)}\)\(=\dfrac{2^{12}.3^{10}.6}{2^{11}.3^{11}.7}=\dfrac{2.2}{7}=\dfrac{4}{7}\)

\(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}+\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{12}}\)

\(=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^{12}.3^6+\left(2^3\right)^4.3^5}-\frac{\left(2^4\right)^3.3^{10}+2^3.3.5.\left(2.3\right)^9}{\left(2^2\right)^6.3^{12}+\left(2.3\right)^{12}}\)

\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6-2^{12}.3^5}-\frac{2^{12}.3^{10}-2^3.3.5.2^9.3^9}{2^{12}.3^{12}+2^{12}.3^{12}}\)

\(=\frac{2^{12}.\left(3^5-3^4\right)}{2^{12}.\left(3^6-3^5\right)}-\frac{2^{12}.3^{10}-2^{12}.3^{10}.5}{2^{12}.3^{12}+2^{12}.3^{12}}\)

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

\(=\frac{162}{486}-\frac{2^{12}.3^{10}.\left(-4\right)}{2^{13}.3^{10}.3^2}=\frac{1}{3}-\frac{2^{14}.3^{10}.\left(-1\right)}{2^{13}.3^{10}.9}\)

\(=\frac{1}{3}-\frac{2.1.\left(-1\right)}{1.1.9}=\frac{1}{3}-\frac{2}{9}=\frac{1}{9}\)

eo ôi t làm rồi mà bị xoá :v thôi  t hướng dẫn :v

Tạc TS và MS ra rồi gộp và triệt tiêu :) nếu k lm đc ibx t làm cho :)

Ta có:

\(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}+\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{12}}\)

\(A=\frac{2^{12}.3^5-\left(2^2\right)^6.9^2}{12^6+8^4.3^5}+\frac{16^3.3^9.3+120.3^9.2^9}{4^6.\left(3^2\right)^6+6^{12}}\)

\(A=\frac{2^{12}.3^5-2^{12}.9^2}{3^6.4^6+8^4.3^5}+\frac{3^9.\left[16^3.3+120.2^9\right]}{4^6.3^{12}+3^{12}.2^{12}}\)

\(A=\frac{2^{12}.162}{3^5.3.4^6+3^5.8^4}+\frac{3^9.73728}{3^{12}.\left(4^6+2^{12}\right)}\)

\(A=\frac{2^{12}.162}{3^5.\left(3.4^6+8^4\right)}+\frac{3^9.73728}{3^{12}.8192}\)

         Đến đây bn tự làm nha hoa mắt lắm ròi

\(\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}=\frac{2^{12}.3^{10}+2^3.3.5.2^9.3^9}{2^{12}.3^{12}+3^{11}.2^{11}}=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{11}.3^{11}\left(2.3+1\right)}=\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}\left(2.3+1\right)}=\frac{2.6}{3.7}=\frac{12}{21}=\frac{4}{7}\)

\(\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}=\frac{4}{7}\)

ngọc diệp làm thế thì ai hiểu

\(y=\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}\)

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

\(y=\frac{2^9.3^9\left(2^3.3+120\right)}{2^{11}.3^{11}\left(2.3+1\right)}\)

\(y=\frac{6^9\left(2^3.3+120\right)}{6^{11}.7}\)

\(y=\frac{2^3.3+120}{6^2.7}\)

\(y=\frac{144}{252}\)

\(y=\frac{4}{7}\)

Lời giải:

Gọi biểu thức là $A$.

\(A=\frac{(2^4)^3.3^{10}+2^3.3.5.2^9.3^9}{2^{12}.3^{12}+2^{11}.3^{11}}\\ =\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{11}.3^{11}(2.3+1)}\\ =\frac{2^{12}.3^{10}(1+5)}{7.2^{11}.3^{11}}=\frac{2^{12}.3^{10}.2.3}{7.2^{11}.3^{11}}\\ =\frac{2^{13}.3^{11}}{7.2^{11}.3^{11}}=\frac{2^2}{7}=\frac{4}{7}\)