Ta có :

\(\left(\frac{16}{25}\right)^{10}=\left(\frac{4^2}{5^2}\right)^{10}=\left(\frac{4}{5}\right)^{2.10}=\left(\frac{4}{5}\right)^{20}\)

\(\left(\frac{3}{7}\right)^{40}=\left(\frac{3}{7}\right)^{2.20}=\left(\frac{3^2}{7^2}\right)^{20}=\left(\frac{9}{49}\right)^{20}\)

Vì 20 = 20 và \(\frac{4}{5}>\frac{9}{49}\)nên \(\left(\frac{4}{5}\right)^{20}>\left(\frac{9}{49}\right)^{20}\)

Vậy \(\left(\frac{16}{25}\right)^{10}>\left(\frac{3}{7}\right)^{40}\)

10²⁰=(10²)¹⁰=100¹⁰>40¹⁰

=> 10²⁰>40¹⁰

**** cho mình

\(5^{200}=\left(5^2\right)^{100}=25^{100}\)

\(3< 25=>3^{100}< 25^{100}=>3^{100}< 5^{200}\)

\(\frac{75^{20}}{45^{10}.25^{15}}=\frac{25^{20}.3^{20}}{3^{10}.3^{10}.5^{10}.25^{15}}=\frac{25^{20}}{25^5.25^{15}}=1\)

\(=>75^{20}=45^{10}.25^{15}\left(dpcm\right)\)

P/S:nếu a=b=>a:b=1 mk làm theo cách đó cho nhanh mà bn ghi sai đề r

a)      Ta có: \(\frac{2}{{ - 5}} = \frac{{ - 16}}{{40}}\) và \(\frac{{ - 3}}{8} = \frac{{ - 15}}{{40}}\)

Do \(\frac{{ - 16}}{{40}} < \frac{{ - 15}}{{40}}\,\, \Rightarrow \,\frac{2}{{ - 5}} < \frac{{ - 3}}{8}\).

b)      Ta có: \( - 0,85 = \frac{{ - 85}}{{100}} = \frac{{ - 17}}{{20}}\). Vậy \( - 0,85\)=\(\frac{{ - 17}}{{20}}\).

c)      Ta có: \(\frac{{37}}{{ - 25}} = \frac{{ - 296}}{{200}}\)  

Do  \(\frac{{ - 137}}{{200}} > \frac{{ - 296}}{{200}}\) nên \(\frac{{ - 137}}{{200}}\) > \(\frac{{37}}{{ - 25}}\) .

d)      Ta có: \( - 1\frac{3}{{10}}=\frac{-13}{10}\) ;

\(-\left( {\frac{{ - 13}}{{ - 10}}} \right) = \frac{{-13}}{{10}}\).

Vậy \(- 1\frac{3}{{10}} =-(\frac{{-13}}{{-10}})\,\).

ta có: \(\left(\frac{16}{25}\right)^{10}=\left[\left(\frac{4}{5}\right)^2\right]^{10}=\left(\frac{4}{5}\right)^{20}\)

\(\left(\frac{3}{7}\right)^{40}=\left[\left(\frac{3}{7}\right)^2\right]^{20}=\left(\frac{9}{49}\right)^{20}\)

mà \(\frac{4}{5}>\frac{9}{49}\)

\(\Rightarrow\left(\frac{4}{5}\right)^{20}>\left(\frac{9}{49}\right)^{20}\)

\(\Rightarrow\left(\frac{16}{25}\right)^{10}>\left(\frac{3}{7}\right)^{40}\)

Đầu tiên ta so sánh 16/25 với 3/7 bằng phương pháp quy đồng mẫu ta được 112/175 > 75/175

Mà với các phân số có tử nhỏ hơn mẫu thì càng lũy thừa >1, phân số càng nhỏ đi. 3/7 có số mũ vượt trội nên khỏi bàn cãi, (3/7)^40 nhỏ hơn rất nhiều.

\(\left(\dfrac{1}{10}\right)^{15}=\left[\left(\dfrac{1}{10}\right)^3\right]^5=\left(\dfrac{1}{1000}\right)^5=\left(\dfrac{10}{10000}\right)^5\)

\(\left(\dfrac{3}{10}\right)^{20}=\left[\left(\dfrac{3}{10}\right)^4\right]^5=\left(\dfrac{81}{10000}\right)^5\)

 \(\dfrac{10}{10000}< \dfrac{81}{10000}\)

\(\Rightarrow\left(\dfrac{10}{10000}\right)^5< \left(\dfrac{81}{10000}\right)^5\)

\(\Rightarrow\left(\dfrac{1}{10}\right)^{15}< \left(\dfrac{3}{10}\right)^{20}\)

Ta có:

\(\left(\dfrac{1}{10}\right)^{15}=\left[\left(\dfrac{1}{10}\right)^3\right]^5=\left(\dfrac{1}{1000}\right)^5\)

\(\left(\dfrac{3}{10}\right)^{20}=\left[\left(\dfrac{3}{10}\right)^4\right]^5=\left(\dfrac{81}{10000}\right)^5\)

Ta thấy: \(\dfrac{1}{1000}< \dfrac{81}{10000}\)

\(\Rightarrow\left(\dfrac{1}{1000}\right)^5< \left(\dfrac{81}{10000}\right)^5\)

\(\Rightarrow\left(\dfrac{1}{10}\right)^{15}< \left(\dfrac{3}{10}\right)^{20}\)

Ta có:

\(\left(\dfrac{1}{10}\right)^{15}=\left(\left(\dfrac{1}{10}\right)^3\right)^5=\left(\dfrac{1}{1000}\right)^5\)

\(\left(\dfrac{3}{10}\right)^{20}=\left(\left(\dfrac{3}{10}\right)^4\right)^5=\left(\dfrac{81}{10000}\right)^5\)

Ta có: \(\left(\dfrac{1}{10}\right)^{15}=\left(\dfrac{1}{10}^3\right)^5=\left(\dfrac{1}{1000}\right)^5\)

\(\left(\dfrac{3}{10}\right)^{20}=\left(\dfrac{3}{10}^4\right)^5=\left(\dfrac{3}{10000}\right)^5\)

Vì \(\dfrac{1}{1000}>\dfrac{3}{10000}\) nên \(\left(\dfrac{1}{10}\right)^{15}>\left(\dfrac{3}{10}\right)^{20}\)