a)\(\left[\frac{3}{7}\cdot\frac{4}{15}+\frac{1}{3}\cdot\left(9^{15}\right)\right]^0\cdot\frac{1}{3}\cdot\frac{6^8}{12^4}\)

\(=1\cdot\frac{1}{3}\cdot\frac{6^4\cdot6^4}{12^4}=\frac{1}{3}\cdot\frac{36^4}{12^4}=\frac{1}{3}\cdot81=27\)

a) \(\frac{10^4.81-16.15^2}{\left(-8\right)^4.3^{12}+6^{11}}=\frac{10^4.3^4-4^2.15^2}{8^4.3^{12}+6^{11}}=\frac{30^4-60^2}{2^{12}.3^{12}+6^{11}}=\frac{\left(30^2\right)^2-60^2}{6^{12}+6^{11}}\)

\(=\frac{900^2-60^2}{6^{11}.\left(6+1\right)}=\frac{60^2.\left(15^2-1\right)}{6^{11}.7}=\frac{60^2.224}{6^{11}.7}=\frac{2^9.3^2.5^2.7}{2^{11}.3^{11}.7}=\frac{5^2}{2^2.3^9}=\frac{25}{78732}\)

180 nha

a: \(A=\dfrac{2^{12}\cdot3^{10}+2^3\cdot2^9\cdot3^9\cdot3\cdot5}{2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}\)

\(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{11}\cdot3^{11}\cdot7}\)

\(=\dfrac{2^{12}\cdot3^{10}\cdot6}{2^{11}\cdot3^{11}\cdot7}=\dfrac{2}{3}\cdot\dfrac{6}{7}=\dfrac{12}{21}=\dfrac{4}{7}\)

b: \(B=\left(\dfrac{12}{105}+\dfrac{9^{15}}{3}\right)\cdot\dfrac{1}{3}\cdot\dfrac{6^8}{6^4\cdot2^4}\)

\(=\dfrac{12+35\cdot9^{15}}{105}\cdot\dfrac{1}{3}\cdot3^4\)

\(=\dfrac{12+35\cdot9^{15}}{105}\cdot3^3=\dfrac{9\left(12+35\cdot9^{15}\right)}{35}\)

\(\frac{10^4.81-16.15^2}{4^4.675}\)

= \(\frac{2^4.5^4.3^4-2^4.3^2.5^2}{4^4.3^3.5^2}\)

= \(\frac{2^4.5^2.3^4-2^4.3^2.5^2}{2^8.3^3.5^2}\)= \(\frac{2^4.5^2.3^2\left(3^2-1\right)}{2^8.3^3.5^2}\)= \(\frac{8}{2^4.3}=\frac{1}{6}\)

mấy bài khó còn làm dc ko lẽ ko làm dc bài này hay quân kiến thức cũ vậy bn

\(\frac{10^4.81-16.15^2}{4^4.675}=\frac{2^4.5^4.3^4-2^4.3^2.5^2}{2^8.3^3.5^2}\)

\(=\frac{2^4.3^2.5^2.\left(5^2.3^2-1\right)}{2^8.3^3.5^2}\)

\(=\frac{25.9-1}{2^4.3}=\frac{225-1}{16.3}=\frac{224}{48}=\frac{14}{3}\)

qwertyuiopasdfgggggghjkllzxcvbnmm,.//234567890-=`

a. - \(\frac{-1}{2}\)

b. \(\frac{238}{3}\)

Kotori Minami bạn có thể giải chi tiết ra duoc ko