\(A=\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\frac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}=\frac{2^{10}.3^8-2^{10}.3^8.3}{2^{10}.3^8+2^{10}.3^8.5}\)

                                   \(=\frac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8.\left(1+5\right)}=-\frac{2}{6}=-\frac{1}{3}\)

\(B=\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\frac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}=\frac{2^{11}.2.3^{10}.\left(1+5\right)}{2^{11}.3^{10}.3.\left(6-1\right)}=\frac{12}{15}=\frac{4}{5}\)

\(A=\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\frac{2^{10}.3^8-2^2.3^9}{2^{10}.3^8-\left(-\left(2^2.3^8.5\right)\right)}=\frac{2^2.3^9}{-\left(2^2.3^8.5\right)}=-\frac{3}{5}\)

thanks bn Phạm Ngọc Thạch nhiều

\(A=\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)
\(=\dfrac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}\)
\(=\dfrac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)
\(=\dfrac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}\)
\(=-\dfrac{2}{6}=-\dfrac{1}{3}\)

\(=\dfrac{\left(2^2\right)^5.\left(3^2\right)^4-2.3^9.2^9}{2^{10}.3^8+2^8.3^8.2^2.5}=\dfrac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}=\dfrac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}=\dfrac{-2}{6}=-\dfrac{1}{3}\)

a: \(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{-2}{6}=\dfrac{-1}{3}\)

b: \(=\dfrac{5^{16}\cdot3^{21}}{5^{15}\cdot3^{22}}=\dfrac{5}{3}\)

a: \(A=\dfrac{5^2\cdot3^{11}\cdot2^{11}\cdot2^8+3^2\cdot2^2\cdot2^{12}\cdot3^6\cdot3^2\cdot5^2}{2\cdot2^{12}\cdot3^{12}\cdot2^4\cdot5^4-3^8\cdot960^3}\)

\(=\dfrac{5^2\cdot3^{11}\cdot2^{19}+3^{10}\cdot2^{14}\cdot5^2}{2^{17}\cdot3^{12}\cdot5^4-3^{11}\cdot2^{18}\cdot5^3}\)

\(=\dfrac{5^2\cdot2^{14}\cdot3^{10}\left(3\cdot2^5+1\right)}{2^{17}\cdot3^{11}\cdot5^3\left(3\cdot5-2\right)}=\dfrac{1}{5}\cdot\dfrac{1}{8}\cdot\dfrac{1}{10}\cdot\dfrac{97}{13}=\dfrac{97}{5200}\)

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

\(=\dfrac{2^{12}\cdot3^{10}\cdot\left(1+5\right)}{2^{11}\cdot3^{11}\left(2\cdot3-1\right)}=\dfrac{2}{3}\cdot\dfrac{6}{5}=\dfrac{12}{15}=\dfrac{4}{5}\)

\(a,A=\left(3\dfrac{5}{6}-1\dfrac{1}{3}\right)\left(3\dfrac{4}{15}-2\dfrac{3}{5}\right)\)
\(\Leftrightarrow A=\left(3+\dfrac{5}{6}-1+\dfrac{1}{3}\right)\left(3+\dfrac{4}{15}-2+\dfrac{3}{5}\right)\)
\(\Leftrightarrow A=\left[\left(3-1\right)+\left(\dfrac{5}{6}+\dfrac{1}{3}\right)\right]+\left[\left(3-2\right)+\left(\dfrac{4}{15}+\dfrac{3}{5}\right)\right]\)
\(\Leftrightarrow A=\left[2+\left(\dfrac{5}{6}+\dfrac{2}{6}\right)\right]+\left[1+\left(\dfrac{4}{15}+\dfrac{9}{15}\right)\right]\)
\(\Leftrightarrow A=\left(2+\dfrac{7}{6}\right)+\left(1+\dfrac{13}{15}\right)\)
\(\Leftrightarrow A=\left(2+1+\dfrac{1}{6}\right)+\left(1+\dfrac{13}{15}\right)\)
\(\Leftrightarrow A=3\dfrac{1}{6}+1\dfrac{13}{15}\)
Vậy...

\(b,B=\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
\(\Leftrightarrow B=\dfrac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^9.\left(2^3.3.5\right)}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}\)
\(\Leftrightarrow B=\dfrac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(\Leftrightarrow B=\dfrac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(\Leftrightarrow B=\dfrac{\left(2^{10}.3^{10}\right)\left(1+5\right)}{\left(2^{11}.3^{11}\right)\left(2.3-1\right)}\)
\(\Leftrightarrow B=\dfrac{6}{\left(2.3\right).5}\)
\(\Leftrightarrow B=\dfrac{6}{6.5}\)
\(\Leftrightarrow B=\dfrac{1}{5}\)
Vậy....

Bài 1:

\(a.5^5-5^4+5^3\)

\(=5^3.5^2-5^3.5+5^3.1\)

\(=5^3\left(5^2-5+1\right)\)

\(=5^3.21\)

\(=5^3.3.7⋮7\)

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Bài 2:

\(a.32< 2^n< 128\)

\(\Rightarrow2^5< 2^n< 2^7\)

\(\Rightarrow n=2\)

\(b.9.27\le3^n\le243\)

\(\Rightarrow3^2.3^3\le3^n\le3^5\)

\(\Rightarrow3^5\le3^n\le3^5\)

\(\Rightarrow n=5\)

Bài 3

\(a.\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)

\(=\dfrac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8+\left(2.3\right)^8.2^2.5}\)

\(=\dfrac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)

\(=\dfrac{2^{10}.3^8.\left(1-3\right)}{2^{10}.3^8.\left(1+5\right)}\)

\(=\dfrac{2^{10}.3^8.\left(-2\right)}{2^{10}.3^8.6}\)

\(=\dfrac{-2}{6}\)

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

Câu b bạn tự làm nha
đưa về lũy thừa có cơ số là 2,3 như câu a rồi rút gọn

\(A=\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)

\(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+3^8\cdot2^{10}\cdot5}\)

\(=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^{10}\cdot3^8\cdot\left(1+5\right)}\)

\(=-\dfrac{2}{6}=-\dfrac{1}{3}\)

a) \(\dfrac{27^3\cdot11+9^5\cdot5}{3^9\cdot2^4}\)

\(=\dfrac{3^9\cdot11+3^{10}\cdot5}{3^9\cdot2^4}\)

\(=\dfrac{3^9\cdot\left(11+3\cdot5\right)}{3^9\cdot2^4}\)

\(=\dfrac{11+15}{16}\)

\(=\dfrac{26}{16}\)

\(=\dfrac{13}{8}\)

b) \(\dfrac{5^8+2^2\cdot25^4+2^3\cdot125^3-15^4\cdot5^4}{4^2\cdot625^2}\)

\(=\dfrac{5^8+2^2\cdot5^8+2^3\cdot5^9-3^4\cdot5^4\cdot5^4}{2^4\cdot5^8}\)

\(=\dfrac{5^8\cdot\left(1+2^2+2^3\cdot5-3^4\right)}{5^8\cdot2^4}\)

\(=\dfrac{1+4+40-81}{16}\)

\(=\dfrac{-36}{16}\)

\(=\dfrac{-9}{4}\)

c) \(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)

\(=\dfrac{2^{10}\cdot3^8-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}\)

\(=\dfrac{2^{10}\cdot3^8\cdot\left(1-3\right)}{2^{10}\cdot3^8\cdot\left(1+5\right)}\)

\(=\dfrac{-2}{6}\)

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

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

Giải giúp em với mn

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