\(n=\frac{6^{6+}6^3\times3^3+3^6}{-73}=\frac{2^6\times3^6+3^6\times2^3+3^6}{-73}=\frac{3^6\times\left(2^6+2^3+1\right)}{-73}=\frac{3^6\times73}{-73}=\left(-3\right)^6=3^6\)

Bài 1,A=\(\frac{644}{193}\)

giải thích cách làm nựa nha bạn

644/193 giải theo tư duy thôi

tả nời

B = 1

-------------------------------------------------------

= \(\frac{1}{6}--\frac{10}{3}\)[1/6 - (-10/3)]

= \(\frac{7}{2}=3,5\)

Không pải vế trước đó nữa cơ vế sau thì ai mà chả làm đc

\(\dfrac{4^0\cdot9^3-6^9\cdot1200}{8^4\cdot3^{12}+6^{11}}\)

= \(\dfrac{1\cdot\left(3^2\right)^3-\left(3\cdot2\right)^9\cdot\left(5^2\cdot2^4\cdot3\right)}{\left(2^3\right)^4\cdot3^{12}+6^{11}}\)

= \(\dfrac{3^6-3^9\cdot2^9\cdot5^2\cdot2^4\cdot3}{2^{12}\cdot3^{12}+\left(2\cdot3\right)^{11}}\)

= \(\dfrac{3^6-3^{10}\cdot2^{13}\cdot5^2}{2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}\)

như lời đã nói thì bn làm tiếp>:)))

\(\frac{2^{12}.3^5-\left(2^2\right)^6.3^6}{2^{12}.\left(3^2\right)^3+\left(2^3\right)^4.3^3}\)

\(\frac{2^{12}.3^5.\left(1-3^{ }\right)}{2^{12}.3^3.\left(3^3-1\right)}\)

\(\frac{2^{12}.3^5.\left(-2\right)}{2^{12}.3^3.8}\)

\(\frac{3^2.\left(-1\right)}{4}\)

\(\frac{-9}{4}\)

VẬy.......................

nhớ tk cho mình nha

\(\frac{3^{17}\cdot81^{11}}{27^{10}\cdot9^{15}}\)

\(=\frac{3^{17}\cdot\left(3^4\right)^{11}}{\left(3^3\right)^{10}\cdot\left(3^2\right)^{15}}\)

\(=\frac{3^{17}\cdot3^{44}}{3^{30}\cdot3^{30}}\)

\(=\frac{3^{61}}{3^{60}}\)

\(=3\)

\(\frac{9^2\cdot2^{11}}{16^2\cdot6^3}\)

\(=\frac{\left(3^2\right)^2\cdot2^{11}}{\left(2^4\right)^2\cdot\left(2\cdot3\right)^3}\)

\(=\frac{3^4\cdot2^{11}}{2^8\cdot2^3\cdot3^3}\)

\(=\frac{3^4\cdot2^{11}}{2^{11}\cdot3^3}\)

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

\(=3\)

\(\frac{2^{10}\cdot3^{31}+2^{40}\cdot3^6}{2^{11}\cdot3^{31}+2^{41}\cdot3^6}\)

\(=\frac{2^{10}\cdot3^{31}+2^{40}\cdot3^6}{2\cdot\left(2^{10}\cdot3^{31}+2^{40}\cdot3^6\right)}\)

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