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

mk ko viết lại đề

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

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

\(=\frac{2}{3.4}+\frac{2^{12}.3^{10}.6}{2.2^{12}.3^{12}}=\frac{1}{6}+\frac{1}{3}=\frac{1}{2}\)

Vậy A= \(\frac{1}{2}\)

\(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)

Phân tích từ số:

\(\frac{2^{12}.3^5-4^2.4^4.3^4}{2^{12}.3.3^5+4^2.4^4.3.3^4}=\frac{1}{6}\)

\(\frac{5^9.5.7^3-5^9.5.7^3.7}{5^9.7^3+5^9.2^3.7^3}=\frac{-10}{3}\)

Sau khi rút gọn là:

\(\frac{1}{6}-\left(-\frac{10}{3}\right)=\frac{1}{6}+\frac{10}{3}=\frac{7}{2}\)

bố mầy đéo hiểu cái éo gì

12fgergtefe

Gợi ý :  Phân tích hết ra thành tích các thừa số nguyên tố rồi đặt cái chung ra ngoài

-> rút gọn

-> kết quả

P/S : bài này cx ko dài lắm nhưg lười  ^^

ko được nói tục

Ta có \(A=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^{12}.3^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^3-\left(5^2\right)^5.\left(7^2\right)^2}{\left(5^3\right)^3.7^3+5^9.7^3.2^3}\)

\(A=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.7^3.2^3}\)

\(=\frac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^3\left(1-7\right)}{5^9.7^3\left(1+2^3\right)}\)

\(=\frac{2}{12}-\frac{5\left(-6\right)}{9}\)\(=\frac{1}{6}+\frac{30}{9}=\frac{7}{2}\)

Vậy A=7/2

A = \(\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}\)- \(\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.7^6}\)

    = \(\frac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^3.\left(1-7\right)}{5^9.7^3.\left(1+2^3\right)}\) = \(\frac{2}{3.4}-\frac{5.\left(-6\right)}{9}\)= \(\frac{1}{6}-\frac{-10}{3}\)= 1/6 + 10/3 = 7/2

7/2 la dung

\(A=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6-2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3-5^9.7^6}\)

\(=\frac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^3.\left(1-7\right)}{5^9.7^3.\left(1+2^3\right)}=\frac{2}{3.4}-\frac{5.\left(-6\right)}{9}=\frac{1}{6}-\frac{-10}{3}=\frac{1}{6}+\frac{10}{3}=\frac{7}{2}\)