a) \(k=\frac{2^{11}.9^2}{3^5.16^2}=\frac{2^{11}.\left(3^2\right)^2}{3^5.\left(2^4\right)^2}=\frac{2^{11}.3^4}{3^5.2^8}=\frac{8.1}{3.1}=\frac{8}{3}\)

b)  \(N=\frac{9^3.27^2}{6^2.3^{10}}=\frac{\left(3^2\right)^3.\left(3^3\right)^2}{\left(2.3\right)^2.3^{10}}=\frac{3^6.3^6}{2^2.3^2.3^{10}}=\frac{3^{12}}{4.3^{12}}=\frac{1}{4}\)

c)\(P=\frac{\left(3^3\right)^{25}.5^3.\left(2^3\right)^4}{\left(5^2\right)^2.\left(3^4\right)^{11}.2^{11}}=\frac{3^{75}.5^3.2^{12}}{5^4.3^{44}.2^{11}}=\frac{3^{31}.1.2}{5.1.1}=\frac{3^{31}.2}{5}=....\)

a) \(K=\frac{2^{11}\cdot9^2}{3^5\cdot16^2}=\frac{2^{11}\cdot3^4}{3^5\cdot2^8}=\frac{2^3}{3}=\frac{8}{3}\)

b) \(N=\frac{9^3\cdot27^2}{6^2\cdot3^{10}}=\frac{3^6\cdot3^6}{2^2\cdot3^2\cdot3^{10}}=\frac{1}{4}\)

c) \(P=\frac{27^{15}\cdot5^3\cdot8^4}{25^2\cdot81^{11}\cdot2^{11}}=\frac{3^{45}\cdot5^3\cdot2^{12}}{5^4\cdot3^{44}\cdot2^{11}}=\frac{3\cdot2}{5}=\frac{6}{5}\)

\(\dfrac{2^{11}.9^2}{3^5.16^2}\)

= \(\dfrac{2^{11}.3^4}{3^5.2^8}\)

= \(\dfrac{2^3}{3}\)= \(\dfrac{8}{3}\)= 2, (6)

a)338800/539000=22/35

b)60/20=1/3

Sửa đề + làm bài \(M=\frac{\frac{2}{3}-\frac{2}{5}+\frac{2}{9}+\frac{2}{13}}{\frac{11}{3}-\frac{11}{5}+\frac{11}{9}+\frac{11}{13}}=\frac{2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{9}+\frac{1}{13}\right)}{11\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{9}+\frac{1}{13}\right)}=\frac{2}{11}\)

Có 1 chỗ bạn ghi sai đề phải là \(\frac{11}{13}\)chứ ko phải \(\frac{11}{3}\)nhé 

\(M=\frac{\frac{2}{3}-\frac{2}{5}+\frac{2}{9}+\frac{2}{13}}{\frac{11}{3}-\frac{11}{5}+\frac{11}{9}+\frac{11}{13}}\)

\(=\frac{2.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{9}+\frac{2}{13}\right)}{11.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{9}+\frac{1}{13}\right)}\)

\(=\frac{2}{11}\)

Học tốt

Mình cảm ơn hai bạn Nguyễn Thị Thu Trang với bạn Xyz ạ , nhưng mà trong sách đề ghi là \(\frac{11}{3}\)

\(\frac{24\cdot47-23}{24+47\cdot23}.\frac{3+\frac{3}{7}-\frac{3}{11}+\frac{3}{1001}-\frac{3}{13}}{\frac{9}{1001}-\frac{9}{13}+\frac{9}{7}-\frac{9}{11}+9}\)

\(=\frac{24\cdot\left(24+23\right)-23}{24+\left(24+23\right)\cdot23}\cdot\frac{3\left(1+\frac{1}{7}-\frac{1}{11}+\frac{1}{1001}-\frac{1}{13}\right)}{9\left(\frac{1}{1001}-\frac{1}{13}+\frac{1}{7}-\frac{1}{11}+1\right)}\)

\(=\frac{24^2+24\cdot23-23}{24+24\cdot23+23^2}\cdot\frac{3}{9}\) \(=\frac{24^2+23\cdot\left(24-1\right)}{\left(23+1\right)\cdot24\cdot23^2}\cdot\frac{1}{3}=1\cdot\frac{1}{3}=\frac{1}{3}\)

giúp giùm mình đi

\(=\dfrac{2^{11}\cdot3^6}{3^5\cdot2^8}=2^3\cdot3=24\)