=  2^10*(3^10-3^9)                   

        2^9*3^10

......................................... bn tự giải tiếp đi nhé, mìh bạn rồi

NHỚ CHO MÌH ĐÓ NHA

\(\frac{2^{10}.3^{10}-2^{10}.3^9}{2^9.3^{10}}\)

\(=\frac{2^{10}\left(3^{10}-3^9\right)}{2^9.3^{10}}\)

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

\(=\frac{2.\left(59049-19683\right)}{59049}\)

\(=\frac{2.39366}{59049}\)

\(=\frac{78732}{59049}\)

mình ko biết vì mình mới học lớp 5 thôi

                               đáp số: mới lớp 5 

\(\frac{2^{10}.3^{10}-2^{10}.3^9}{2^9.3^{10}}=\frac{2^{10}\left(3^{10}-3^9\right)}{2^9.3^{10}}\)

\(=\frac{2^{10}.3}{2^9.3^{10}}=\frac{2^9.2.3}{2^9.3^9.3}\)

\(=\frac{2}{3^9}\)

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

tinh song roi rut gon 

rút gọn ta đc kết quả \(\frac{4}{3}\)

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

dòng đầu tiên là 

ta có 2^10

nhé

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

\(\frac{2^{10}.3^{10}-3^{10}.3^9}{2^9.3^{10}}=\frac{3^{10}\left(2^{10}-3^9\right)}{2^9.3^{10}}=\frac{2^{10}-3^9}{2^9}\)

\(\left(-\right)\frac{1989\cdot1990+3970}{1992\cdot1991+3984}=\frac{1989\cdot\left(1990+2\right)}{1992\cdot\left(1991+2\right)}=\frac{1989}{1993}\)

\(\left(-\right)\frac{3^{10}\cdot\left(-5\right)^{21}}{\left(-5\right)^{20}\cdot3^{12}}=\frac{3^{10}\cdot\left(-5\right)^{20}\cdot\left(-5\right)}{3^{10}\cdot\left(-5\right)^{20}\cdot3^2}=-\frac{5}{9}\)

\(\left(-\right)\frac{\left(-11\right)^5\cdot13^7}{11^5\cdot13^8}=\frac{11^5\cdot13^7\cdot\left(-1\right)}{11^5\cdot13^7\cdot13}=-\frac{1}{13}\)

\(\left(-\right)\frac{2^{10}\cdot3^{10}-2^{10}\cdot3^9}{2^9\cdot3^{10}}=\frac{2^{10}\cdot3^9\left(3-1\right)}{2^9\cdot3^{10}}=\frac{2^{11}\cdot3^9}{2^9\cdot3^{10}}=\frac{4}{3}\)

 a) \(\frac{{ - 3}}{8} = \frac{{ - 3.3}}{{8.3}} = \frac{{ - 9}}{{24}}\)

Vì -9 < -5 nên \(\frac{{ - 9}}{{24}} < \frac{{ - 5}}{{24}}\)

Vậy \(\frac{{ - 3}}{8} < \frac{{ - 5}}{{24}}\).

b) Cách 1: \(\frac{{ - 2}}{{ - 5}} = \frac{2}{5}; \frac{3}{{ - 5}} = \frac{-3}{{5}}\)

Vì 2 > -3 nên \(\frac{2}{5} > \frac{-3}{{5}}\)

Vậy \(\frac{{ - 2}}{{ - 5}} > \frac{3}{{ - 5}}\).

Cách 2: \(\frac{{ - 2}}{{ - 5}} = \frac{2}{5} > 0\) mà \(\frac{3}{{ - 5}} < 0\)

\(\Rightarrow\) \(\frac{{ - 2}}{{ - 5}} > \frac{3}{{ - 5}}\).

c) \(\frac{{ - 3}}{{ - 10}} = \frac{3}{{10}} = \frac{{3.2}}{{10.2}} = \frac{6}{{20}}\)

\(\frac{{ - 7}}{{ - 20}} = \frac{7}{{20}}\)

Vì 6 < 7 nên \(\frac{6}{{20}} < \frac{7}{{20}}\) nên \(\frac{{ - 3}}{{ - 10}} < \frac{{ - 7}}{{ - 20}}\).

d) \(\frac{{ - 5}}{4} = \frac{{ - 5.5}}{{4.5}} = \frac{{ - 25}}{{20}}; \frac{{ 23}}{{-20}}=\frac{{-23}}{{20}} \)

Vì -25 < -23 nên \( \frac{{ - 25}}{{20}} < \frac{{-23}}{{20}} \)

Vậy \(\frac{{ - 5}}{4} < \frac{{23}}{{ - 20}}\).