\(=\frac{3^7\cdot\left(2^4\right)^3}{\left(2^2\cdot3\right)^5\cdot\left(3^3\right)^2}\)    

\(=\frac{3^7\cdot2^{12}}{2^{10}\cdot3^5\cdot3^6}\)    

\(=\frac{3^7\cdot2^{12}}{2^{10}\cdot3^{11}}\)   

\(=\frac{2^2}{3^4}\)   

\(=\frac{4}{81}\)

a)\(=\frac{\left(-0,3\right)^7.2^8}{\left(-0,3\right)^7.2^7.\left(-1\right)}\)

\(=\frac{2}{-1}=-2\)

b)\(=\frac{3^7.2^{12}}{2^{10}.3^{11}}\)

\(=\frac{2^2}{3^4}=\frac{4}{81}\)

\(A=\frac{2^{12}.27^3}{6^7.16^2}=\frac{2^{12}.\left(3^3\right)^3}{2^7.3^7.\left(2^4\right)^2}=\frac{2^{12}.3^9}{2^7.3^7.2^8}=\frac{2^{12}.3^9}{2^{15}.3^7}=\frac{1.3^2}{2^3.1}=\frac{9}{8}\)

\(A=\frac{2^{12}\cdot27^3}{6^7\cdot16^2}\)

\(A=\frac{4096\cdot19683}{279936\cdot256}\)

\(A=\frac{80621568}{71663616}\)

\(\dfrac{3}{5}.\dfrac{6}{7}+\dfrac{3}{7}:\dfrac{5}{3}-\dfrac{2}{7}:1\dfrac{2}{3}.\)

\(=\dfrac{3}{5}.\dfrac{6}{7}+\dfrac{3}{7}.\dfrac{3}{5}-\dfrac{2}{7}.\dfrac{3}{5}.\)

\(=\dfrac{3}{5}\left(\dfrac{6}{7}+\dfrac{3}{7}-\dfrac{2}{7}\right).\)

\(=\dfrac{3}{5}.1=\dfrac{3}{5}.\)

Vậy.....

\(\dfrac{3^7.16^3}{12^5.27^2}=\dfrac{3^7.\left(2^4\right)^3}{\left(2^2.3\right)^5.\left(3^3\right)^2}.\)

\(=\dfrac{3^7.2^{12}}{\left(2^2\right)^5.3^5.3^6}=\dfrac{3^7.2^{12}}{2^{10}.3^{11}}.\)

\(=\dfrac{2^2}{3^4}=\dfrac{4}{81}.\)

Vậy.....

a)F=-2

b)G=1536

c)H=229582512

1) \(A=3\sqrt{\dfrac{1}{3}}-\dfrac{5}{2}\sqrt{12}-\sqrt{48}\)

\(=3\cdot\dfrac{\sqrt{1}}{\sqrt{3}}-\dfrac{5\sqrt{12}}{2}-\sqrt{4^2\cdot3}\)

\(=\dfrac{3\cdot1}{\sqrt{3}}-\dfrac{5\cdot2\sqrt{3}}{2}-4\sqrt{3}\)

\(=\sqrt{3}-5\sqrt{3}-4\sqrt{3}\)

\(=-8\sqrt{3}\)

2) \(A=\sqrt{12-4x}\) có nghĩa khi:

\(12-4x\ge0\)

\(\Leftrightarrow4x\le12\)

\(\Leftrightarrow x\le\dfrac{12}{4}\)

\(\Leftrightarrow x\le3\)

3) \(\dfrac{2x-2\sqrt{x}}{x-1}\)

\(=\dfrac{2\sqrt{x}\cdot\sqrt{x}-2\sqrt{x}}{\left(\sqrt{x}\right)^2-1^2}\)

\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{2\sqrt{\text{x}}}{\sqrt{x}+1}\)