Đề bài yêu cầu gì?

sory em học lớp 5 không biết làm nếu biết em đã làm rồi hihihih.....

Có ai trả lời đi

giups mình với mai mình thi rồi!

\(\frac{2^5.9^3}{3^4.2^7}=\frac{\left(3^2\right)^3}{3^4.2^2}=\frac{3^6}{3^4.2^2}=\frac{3^2}{2^2}=\frac{9}{4}\)

quá dễ!!!!!!

a: \(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{-2}{6}=\dfrac{-1}{3}\)

b: \(=\dfrac{5^{16}\cdot3^{21}}{5^{15}\cdot3^{22}}=\dfrac{5}{3}\)

\(A=\dfrac{2}{4.7}-\dfrac{3}{5.9}+\dfrac{2}{7.10}-\dfrac{3}{9.13}+...+\dfrac{2}{301.304}-\dfrac{3}{401.405}\)

\(A=\dfrac{2}{4.7}+\dfrac{2}{7.10}+\dfrac{2}{301.304}...-\left(\dfrac{3}{5.9}+\dfrac{3}{9.13}+...+\dfrac{3}{401.405}\right)\)

\(A=2\left(\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{301.304}\right)...-3\left(\dfrac{1}{5.9}+\dfrac{1}{9.13}+...+\dfrac{1}{401.405}\right)\)

\(A=2\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{301}-\dfrac{1}{304}\right)...-3\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{401}-\dfrac{1}{405}\right)\)

\(A=2\left(\dfrac{1}{4}-\dfrac{1}{304}\right)-3\left(\dfrac{1}{5}-\dfrac{1}{405}\right)\)

\(A=2\left(\dfrac{76}{304}-\dfrac{1}{304}\right)-3\left(\dfrac{81}{5}-\dfrac{1}{405}\right)\)

\(A=2.\dfrac{75}{304}-3.\dfrac{80}{405}=\dfrac{75}{152}-\dfrac{80}{135}=\dfrac{10125-12160}{152.135}=-\dfrac{2035}{152.135}=-\dfrac{407}{4104}\)

\(\left(2^5.9-2^5.7\right)\div2^3-2021^0\)

\(=2^5\left(9-7\right)\div2^3-1\)

\(=2^6\div2^3-1\)

\(=2^3-1\)

\(=7\)

7

a) 5.9 > 0

b) (-3) . (-47) > 15

c) (-3) .(-2) > (-3)

d) (-9) .(-7) > (9)