Ta có:\(\frac{3^{10}\cdot\left(-5\right)\cdot21}{\left(-5\right)^{20}\cdot3^{12}}\)

\(=\frac{3^{10}\cdot\left(-5\right)\cdot21}{\left(-5\right)^{19}\cdot\left(-5\right)\cdot3^{12}}\)

\(=\frac{3\cdot7}{\left(-5\right)^{19}\cdot3^2}\)

\(=\frac{7}{\left(-5\right)^{19}\cdot3}\)

Ảnh hơi mờ 

\(\frac{3^{10}.\left(-5\right)^{21}}{\left(-5\right)^{20}.3^{12}}=\frac{3^{10}.\left(-5\right)^{20}.\left(-5\right)}{\left(-5\right)^{20}.3^{10}.3^2}=\frac{-5}{3^2}=-\frac{5}{9}\)

\(\frac{3^{10}.\left(-5\right)^{20}.\left(-5\right)}{3^{10}.3^2.\left(-5\right)^{20}}=\frac{-5}{3^2}=\frac{-5}{9}\)

T I C K cho mình nha cảm ơn chúc bạn học tốt nha 

\(=\frac{3^{10}.\left(-5\right)^{20}.\left(-5\right)^1}{\left(-5\right)^{20}.3^{10}.3^2}\)\(=\frac{\left(-5\right)^1}{3^2}\)\(=\frac{-5}{9}\)

Hok tốt !

\(\frac{3^{10}.\left(-5\right)^{20}.\left(-5\right)^1}{\left(-5\right)^{20}.3^{10}.3^2}=\frac{-5}{3^2}=\frac{-5}{9}\)

 \(y=\frac{3^{10}.\left(-5\right)^2}{\left(-5\right)^{20}.3^{12}}=\frac{1.1}{5^{18}.3^{10}}=\frac{1}{5^8.5^{10}.3^{10}}=\frac{1}{5^5.15^{10}}\)

Hai câu a) và b) bạn chỉ cần xem số mũ rồi trừ số mũ là xong 

\(c)\) \(\frac{2^{10}.3^{10}-2^{10}.3^9}{2^9.3^{10}}=\frac{2^{10}.3^9\left(3-1\right)}{2^9.3^{10}}=\frac{2^{10}.3^9.2}{2^9.3^{10}}=\frac{2^{11}.3^9}{2^9.3^{10}}=\frac{2^2}{3}=\frac{4}{3}\)

\(d)\) \(\frac{5^{11}.7^{12}+5^{11}.7^{11}}{5^{12}.7^{12}+9.5^{11}.7^{11}}=\frac{5^{11}.7^{11}\left(7+1\right)}{5^{11}.7^{11}\left(5.7+9\right)}=\frac{8}{35+9}=\frac{8}{44}=\frac{2}{11}\)

Chúc bạn học tốt 

Ta có : 21^0.3^10−2^10.3^9/2^9.3^10
=> 2^10.(3^10−3^9)/2^9.3^10
=> 2^10.3/2^9.3^10
=> 2/3^9=2/19683

\(\frac{3^{10}.\left(-5\right)^{21}}{\left(-5\right)^{20}.3^{12}}=\frac{1.\left(-5\right)}{1.3^2}=-\frac{5}{9}\)

Chúc bạn học tốt

Giari thích :

 \(3^{12}:3^{10}=3^{12-10}=3^2=9\)

\(\left(-5\right)^{21}:\left(-5\right)^{20}=\left(-5\right)^{21-20}=\left(-5\right)^1=\left(-5\right)\)

\(\frac{3^{10}.\left(-5\right)^{21}}{\left(-5\right)^{20}.3^{12}}\)

\(=\frac{\left(-5\right)}{3^2}\)

\(=\frac{-5}{9}\)

OK