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\(=\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}\)

\(=\dfrac{5^{10}\cdot7^4\cdot\left(1-7\right)}{5^9\cdot7^3\cdot\left(1+2^3\right)}=\dfrac{5\cdot7\cdot\left(-6\right)}{9}=\dfrac{-2\cdot5\cdot7}{3}=\dfrac{-70}{3}\)

\(A=\dfrac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^6+2^{12}\cdot3^5}-\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}\)

\(=\dfrac{3^4\left(3-1\right)}{3^5\left(3+1\right)}-\dfrac{5^{10}\cdot7^3\left(1-7\right)}{5^9\cdot7^3\cdot9}\)

\(=\dfrac{1}{3\cdot2}-\dfrac{1}{5}\cdot\dfrac{-6}{9}=\dfrac{1}{6}+\dfrac{6}{45}=\dfrac{45+36}{270}=\dfrac{81}{270}=\dfrac{3}{10}\)

25 tháng 5 2022

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

\(=\dfrac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\dfrac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}\)

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

\(=\dfrac{2^{12}.3^4.2}{2^{12}.3^5.4}-\dfrac{5^{10}.7^3.\left(-6\right)}{5^9.7^3.9}\)

\(=\dfrac{1}{6}-\dfrac{-10}{3}\)

\(=\dfrac{7}{2}\)

25 tháng 5 2022

hack não qué

24 tháng 7 2017

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

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

\(=\dfrac{2^{12}(3^5-3^4)}{2^{12}(3^6+3^5)}-\dfrac{5^{10}(7^3-7^4)}{5^9.7^3(1+2^3)}\)

\(=\dfrac{2^{12}.162}{2^{12}.972}-\dfrac{5^{10}(-2058)}{5^9.7^3.9}\)

\(=\dfrac{2^{12}.162}{2^{12}.972}-\dfrac{5^{10}(-2058)}{5^9.7^3.9}\)

\(=\dfrac{162}{972}-\dfrac{5(-2058)}{7^3.9}\)

\(=\dfrac{2.3^4}{2^2.3^5}-\dfrac{5.2.7^3.\left(-3\right)}{7^3.3^2}\)

\(=\dfrac{1}{2.3}-\left(\dfrac{-\left(5.2\right)}{3}\right)\)

\(=\dfrac{1}{6}-\left(\dfrac{-10}{3}\right)\)

\(=\dfrac{7}{2}\)

29 tháng 11 2017

\(A=\dfrac{2^{15}.3^5-4^6.9^2}{\left(2^2.3\right)+8^4.3^5}-\dfrac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)

\(\Leftrightarrow A=\dfrac{2^{15}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^2.3+\left(2^3\right)^4.3^5}-\dfrac{5^{10}.7^3-\left(5^2\right)^5.\left(7^2\right)^2}{\left(5^3\right)^3.7^3+5^9.14^3}\)

\(\Leftrightarrow A=\dfrac{2^{15}.3^5-2^{12}.3^4}{2^2.3+2^{12}.3^5}-\dfrac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.\left(7.2\right)^3}\)

\(\Leftrightarrow A=\dfrac{2^{12}.3^4\left(2^3.3-1\right)}{2^2.3\left(1+2^{10}.3^4\right)}-\dfrac{5^{10}.7^3\left(1-7\right)}{5^9.7^3\left(1+2^3\right)}\)

\(\Leftrightarrow A=\dfrac{2^{10}.3^3\left(2^3.3-1\right)}{\left(1+2^{10}.3^4\right)}-\dfrac{5\left(1-7\right)}{\left(1+2^3\right)}\)

\(\Leftrightarrow A=\dfrac{1024.9\left(8.3-1\right)}{\left(1+1024.81\right)}-\dfrac{5\left(1-7\right)}{\left(1+8\right)}\)

\(\Leftrightarrow A=\dfrac{9216\left(24-1\right)}{\left(1+82944\right)}-\dfrac{5\left(1-7\right)}{\left(1+8\right)}\)

\(\Leftrightarrow A=\dfrac{9216.23}{82945}-\dfrac{5\left(-6\right)}{9}\)

\(\Leftrightarrow A=\dfrac{211968}{82945}+\dfrac{30}{9}\)

\(\Leftrightarrow A=\dfrac{1907712}{746505}+\dfrac{2488350}{746505}\)

\(\Leftrightarrow A=\dfrac{1907712+2488350}{746505}\)

\(\Leftrightarrow A=\dfrac{4396062}{746505}\)

\(A=\dfrac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^6\cdot3^3+2^{12}\cdot3^5}-\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}\)

\(=\dfrac{2^{12}\cdot3^4\cdot\left(3-1\right)}{2^6\cdot3^3\cdot\left(1+2^6\cdot3^2\right)}-\dfrac{5^{10}\cdot7^3\cdot\left(1-7\right)}{5^9\cdot7^3\cdot\left(1+2^3\right)}\)

\(=\dfrac{2^6\cdot3\cdot2}{1+64\cdot9}-\dfrac{5\cdot\left(-6\right)}{9}\)

\(=\dfrac{2^7\cdot3}{1+576}+\dfrac{30}{9}\)

\(=\dfrac{384}{577}+\dfrac{30}{9}=\dfrac{384}{577}+\dfrac{10}{3}=\dfrac{6922}{1731}\)

15 tháng 8 2017

 \(\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6.2^{12}.3^5}\)\(\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.7^3.2^3}\)\(\frac{2^{12}.3^4.\left(3-1\right)}{2^{24}.3^{11}}\)-\(\frac{5^{10}.7^3.\left(1-7\right)}{5^9.7^3.\left(1+8\right)}\)=\(\frac{1}{2^{11}.3^7}\)-\(\frac{-10}{3}\)

15 tháng 8 2017

hình như bạn viết sai đề bài rồi 

3 tháng 10 2019

TA có\(\frac{2^{12}.3^5.4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3}\)

=\(\frac{2^{12}.3^5.2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3}\)

=\(\frac{2^{24}.3^9}{2^{12}.3^5.\left(3+1\right)}-\frac{5^{10}.7^3.\left(1-7\right)}{5^9.7^3}\)

=\(2^{20}.3^4-30\)