BÀI 1

\(\frac{2^7.9^3}{6^5.8^2}=\frac{2^7.\left(3^2\right)^3}{\left(2.3\right)^5.\left(2^3\right)^2}=\frac{2^7.3^6}{2^5.3^5.2^6}=\frac{3}{2^4}=\frac{3}{16}.\)

bài 2

a)           \(\frac{1}{2}-\frac{1}{3}+\frac{1}{12}=\frac{6}{12}-\frac{4}{12}+\frac{1}{12}=\frac{3}{12}=\frac{1}{4}\)

b)          \(\frac{9^9.27^4}{3^8.81^5}=\frac{\left(3^2\right)^9.\left(3^3\right)^4}{3^8.\left(3^4\right)^5}=\frac{3^{18}.3^{12}}{3^8.3^{20}}=\frac{3^{30}}{3^{28}}=3^2=9\)

Study well 

Bài 1: \(\frac{2^7.9^3}{6^5.8^2}=\frac{2^7.3^6}{2^5.3^5.2^6}=\frac{3}{2^4}=\frac{3}{16}\)

Bài 2: 

a)\(\frac{1}{2}-\frac{1}{3}+\frac{1}{12}=\frac{6}{12}-\frac{4}{12}+\frac{1}{12}=\frac{6-4+1}{12}=\frac{1}{4}\)

b)\(\frac{9^9.27^4}{3^8.81^5}=\frac{9^9.3^{12}}{3^8.9^{10}}=\frac{3^4}{9}=\frac{3^4}{3^2}=3^2=9\)

có kq ch bn ?

11/4 giờ = bao nhiêu phút

lop6 đi mà học

\(M=\frac{9^4.27^5.3^6.3^4}{3^8.81^4.234.8^2}=\frac{\left(3^2\right)^4.\left(3^3\right)^5.3^6.3^4}{3^8.\left(3^4\right)^4.3^5.\left(2^3\right)^2}\)

\(M=\frac{3^8.3^{15}.3^6.3^4}{3^{18}.3^{16}.3^5.2^6}=\frac{81}{64}\)

\(N=\frac{4^6.9^5.6^9.120}{8^4.3^{12}-6^{11}}=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)

\(N=\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.\left(2.3-1\right)}=\frac{2.6}{3.5}\)

\(N=\frac{4}{5}\)

-0,26

p/s: bấm máy tính có thể sai chỗ nào đó nên sai không chịu trách nhiệm đâu nhé @@

=\(-\frac{13}{50}\)=-0,26

dễ vậy hỏi làm gì.chịu khó bấm máy cho quen

tối mk làm cho nha, giờ mk đg bận, sorry

\(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)+8^4.3^5}\)

\(\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^2.3+2^4.3^4.3^5}\)

\(\frac{2^{12}.3^5-2^{12}.3^4}{2^2.3+2^4.3^{20}}\)

\(\frac{2^{12}.3^4\left(3-1\right)}{2^2.3\left(1+2^2.3^{19}\right)}\)

\(\frac{2^{10}.3^2.4}{1+2^2.3^{19}}\)

bn tự giải tiếp nha mk phải có việc rồi nếu ko lm đc thì kb với mk tối mk giải cho nha

\(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6}-\frac{5^{10}.7^4-25^5.49^2}{\left(125.7\right)3+5^9.\left(14\right)^3}\)

\(=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{\left(2^2\right)^6.3^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^3-\left(5^2\right)^5.\left(7^2\right)^2}{125^3.7^3+5^9.\left(2.7\right)^3}\)

\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.7^3.2^3}\)

\(=\frac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^3\left(1-7\right)}{5^9.7^3\left(1+8\right)}\)

\(=\frac{2}{3.4}-\frac{5.\left(-6\right)}{9}=\frac{2}{12}-\frac{-30}{9}\)

\(=\frac{1}{6}+\frac{10}{3}=\frac{1}{6}+\frac{20}{6}=\frac{21}{6}=\frac{7}{2}\)

Bạn ơi cho mk hỏi chỗ đoạn kia bạn  lấy 1-7 ở đâu và 1 + 8 ở đâu

tại vì 10⁵.7³-10⁵.7⁴ ko khác j 10⁵.7³.1-10⁵.7³.7 suy ra đc 10⁵.7³.(1-7)

chúc học giỏi

\(\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}=\frac{5.2^{30}.3^{18}-2^2.2^{27}.3^{20}}{5.2^9.2^{19}.3^{19}-7.2^{29}-3^{18}}=\frac{2^{29}.3^{18}\left(5.2-3^2\right)}{2^{28}.3^{18}\left(5.3-7.2\right)}=\frac{2.1}{1}=2\)

\(y=\frac{7.9+14.27+21.36}{21.27+42.81+63.108}=\frac{7.9+2.7.3.9-3.7.4.9}{3.7.3.9+2.3.7.9.9+7.9.4.3.9}=\frac{7.9\left(1+6-12\right)}{3.7.9\left(3+18+36\right)}=\frac{-5}{3.57}=\frac{-5}{171}\)