2^12.3^5-4^6.9^2/(2^2.3)^6
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A=212x35-46x92:(22x3)6+84x35
=212x35-212x34:212x36+212x35
=212(35-34+35):212x36=212x405:(212x36)=1x\(\dfrac{5}{9}\)=\(\dfrac{5}{9}\).


Ta có: \(\frac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6}\)
\(=\frac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^6}=\frac{2^{12}\cdot3^4\left(3-1\right)}{2^{12}\cdot3^6}\)
\(=\frac{3^4\cdot2}{3^6}=\frac{2}{3^2}=\frac29\)
Ta có: \(\frac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\)
\(=\frac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}\)
\(=\frac{5^{10}\cdot7^3\left(1-7\right)}{5^9\cdot7^3\left(1+2^3\right)}=5\cdot\frac{-6}{9}=5\cdot\frac{-2}{3}=-\frac{10}{3}\)
Ta có: \(A=\frac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6}-\frac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\)
\(=\frac29+\frac{10}{3}=\frac29+\frac{30}{9}=\frac{32}{9}\)

a) 81^11.3^17/27^10.9^15
=(9^2)^11.3^17/(3^3)^10.9^15
=3^44.3^17/3^30.3^30
=3^61/3^60
=3
b) A =( (2^12.3^5 - 2^12.3^4)/ (2^12.3^6 + 2^12.3^5) ) - ((5^10.7^3 - 5^10.7^4)/(5^9.7^3 + 5^9.2^3.7^3))
=(2^12.3^4(3-1))/2^12.3^5(3+1) - 5^10.7^3(1-7)/5^9.7^3(1+8)
=2/12- (-30/9)=1/6 + 10/3 = 7/2

\(\dfrac{2^{12}.3^{5}-4^{6}.9^{2}}{(2^{2}.3)^{6}}\)
\(=\dfrac{2^{12}.3^{5}-2^{12}.3^{4}}{2^{12}.3^{6}}\)
\(=\dfrac{2^{12}.3^{4}(3-1)}{2^{12}.3^{4}.3^{2}}\)
\(=\dfrac{2}{9}\)