\(\dfrac{45^{10}\cdot5^{20}}{75^{15}}=\dfrac{\left(3^2\cdot5\right)^{10}\cdot5^{20}}{\left(3\cdot5^2\right)^{15}}=\dfrac{3^{20}\cdot5^{10}\cdot5^{20}}{3^{15}\cdot5^{30}}=3^5=243\\ \dfrac{6^6+6^3+3^3+3^6}{-73}=\dfrac{46656+216+27+729}{-73}=-\dfrac{47628}{73}\\ \dfrac{27^7+3^{15}}{9^9-27}=\dfrac{\left(3^3\right)^7+3^{15}}{\left(3^2\right)^9-3^3}=\dfrac{3^{21}+3^{15}}{3^{18}-3^3}=\dfrac{3^{15}\left(3^6+1\right)}{3^3\left(3^{15}-1\right)}=\dfrac{3^5\cdot730}{3^{15}-1}\\ \dfrac{8^{20}+4^{20}}{4^{25}+64^5}=\dfrac{\left(2^3\right)^{20}+\left(2^2\right)^{20}}{\left(2^2\right)^{25}+\left(2^6\right)^5}=\dfrac{2^{60}+2^{40}}{2^{50}+2^{30}}=\dfrac{2^{40}\left(2^{20}+1\right)}{2^{30}\left(2^{20}+1\right)}=2^{10}=1024\)

\(\dfrac{3}{5}+\dfrac{-1}{25}-\dfrac{8}{20}\)

\(=\dfrac{15}{25}+\dfrac{-1}{25}-\dfrac{8}{20}\)

\(=\dfrac{14}{25}-\dfrac{2}{5}\)

\(=\dfrac{14}{25}-\dfrac{10}{25}\)

\(=\dfrac{4}{25}\)

\(=\dfrac{15}{100}+\dfrac{-4}{100}-\dfrac{40}{100}\)

\(=\dfrac{-29}{100}\)

Em đăng vào môn Toán nha

e: \(=\dfrac{5^{30}\cdot3^{20}}{3^{15}\cdot5^{30}}=3^5=243\)

`(13+x)/20 = 3/4`

`(13+x) xx4=3xx20`

`(13+x)xx4=60`

`13+x=60:4`

`13+x=15`

`x=15-13`

`x=2`

__

`(23-x)/25 =4/5`

`(23-x)xx5=4xx25`

`(23-x)xx5=100`

`23-x=100:5`

`23-x=20`

`x=23-20`

`x=3`

\(1,25:\dfrac{15}{20}+\left(25\%-\dfrac{5}{6}\right):4\dfrac{2}{3}\)

\(=\dfrac{5}{4}.\dfrac{4}{3}+\left(\dfrac{1}{4}-\dfrac{5}{6}\right).\dfrac{3}{14}\)

\(=\dfrac{5}{3}+\dfrac{7}{12}.\dfrac{3}{14}\)

\(=\dfrac{5}{3}+-\dfrac{1}{8}\)

\(=\dfrac{37}{24}\)

\(1,25:\dfrac{15}{20}+\left(25\%-\dfrac{5}{6}\right):4\dfrac{2}{3}\)

\(=\dfrac{5}{4}:\dfrac{3}{4}+\left(\dfrac{1}{4}-\dfrac{5}{6}\right):\dfrac{14}{3}\)

\(=\dfrac{5}{4}.\dfrac{4}{3}+\left(\dfrac{-7}{12}\right).\dfrac{3}{14}\)

= \(\dfrac{5}{3}-\dfrac{1}{8}=\dfrac{37}{24}\)

a) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2003.2004}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2003}-\dfrac{1}{2004}=1-\dfrac{1}{2004}=\dfrac{2003}{2004}\)b)Đặt  \(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2003.2005}\)

\(\Rightarrow2A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2003.2005}=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2003}-\dfrac{1}{2005}=1-\dfrac{1}{2005}=\dfrac{2004}{2005}\)\(\Rightarrow A=\dfrac{1002}{2005}\)

a: Ta có: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2003\cdot2004}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2003}-\dfrac{1}{2004}\)

\(=\dfrac{2003}{2004}\)

b: Ta có: \(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{2003\cdot2005}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{2003\cdot2005}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2003}-\dfrac{1}{2005}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2004}{2005}=\dfrac{1002}{2005}\)

\(a,\dfrac{20}{30}=\dfrac{30}{45}v\text{ì}20.45=30.30=900\\ b,\dfrac{-25}{35}=\dfrac{-55}{77}v\text{ì}:\left(-25\right).77=35.\left(-55\right)=-1925\)

Chưa ghi rõ đề \(\Rightarrow\) không làm được