\(\frac{27^2.8^5}{6^6.32^3}=\frac{\left(3^3\right)^2.\left(2^3\right)^5}{2^6.3^6.\left(2^5\right)^3}\)

               \(=\frac{3^6.2^{15}}{2^6.3^6.2^{15}}\)

                \(=\frac{1}{2^6}\)

\(\frac{27^2.8^5}{6^6.32^3}=\frac{\left(3^3\right)^2.\left(2^3\right)^5}{\left(2.3\right)^6.\left(2^5\right)^3}=\frac{3^6.2^{15}}{2^6.3^6.2^{15}}=\frac{1}{2^6}=\frac{1}{64}\)

cảm ơn nhé do not ask why

\(\dfrac{\left(3^3\right)^2.\left(2^3\right)^5}{2^6.3^6.\left(2^4\right)^3}=\dfrac{3^6.2^{15}}{2^{18}.3^6}=\dfrac{1}{2^{13}}=2^{-13}\)

\(2\sqrt{48}-3\sqrt{75}+5\sqrt{3}\)

\(=2\sqrt{16.3}-3\sqrt{25.3}+5\sqrt{3}\)

\(=2\sqrt{4^2.3}-3\sqrt{5^2.3}+5\sqrt{3}\)

\(=2.4\sqrt{3}-3.5\sqrt{3}+5\sqrt{3}\)

\(=8\sqrt{3}-15\sqrt{3}+5\sqrt{3}\)

\(=\left(8-15+5\right).\sqrt{3}\)

\(=-2\sqrt{3}\)

\(-\dfrac{1}{2}x+6< 0\Leftrightarrow-\dfrac{1}{2}x< -6\Leftrightarrow\cdot\dfrac{1}{2}x>6\Leftrightarrow x>12\)

(sai thì thoi nha)

\(-\dfrac{1}{2}x+6< 0\)

\(\Leftrightarrow-\dfrac{1}{2}x< -6\)

\(\Leftrightarrow x>\left(-6\right):\left(-\dfrac{1}{2}\right)\)

\(\Leftrightarrow x>12\)

--> Chọn A

a: \(=\dfrac{-3^4\cdot2^8}{2^2\cdot2^2\cdot3^2}=-3^2\cdot2^2=-6^2=-36\)

b: \(=\dfrac{3^6\cdot2^{15}}{3^6\cdot2^6\cdot2^{15}}=\dfrac{1}{2^6}=\dfrac{1}{64}\)

c: \(=\left(\dfrac{0.8}{0.4}\right)^5\cdot\dfrac{1}{0.4}=2^5\cdot\dfrac{1}{0.4}=\dfrac{32}{0.4}=80\)

a,=1/8*1/16=1/108

b,=(27/9)^3=3^3=27

c,=125^2/25^2*25=5^2*25=25*25=225

789,2 × 27 = 21307,4