\(A=\frac{3^8.\left(2^3\right)^5}{8^3.3^5}\)

\(=\frac{3^5.3^3.8^5}{8^3.3^5}\)

\(=\frac{3^5.8^3.3^3.8^2}{8^3.3^5}\)

\(=3^3.8^2=1728\)

Ta có \(A=\frac{3^8.2^{15}}{8^3.3^5}\)

\(=\frac{3^8.2^{15}}{2^9.3^5}\)

\(=\frac{3^3.2^6}{1}\)

\(=91\)

Vậy....

\(A=\frac{3^8.2^{15}}{\left(2^3\right)^3.3^5}\)

\(A=\frac{3^8.2^{15}}{2^9.3^5}\)

\(A=\frac{3^3.2^6}{1}\)

\(A=1728\)

\(A=\dfrac{12^{15}\cdot3^4-4^5\cdot3^9}{27^3\cdot2^{10}-32^3\cdot3^9}\\ =\dfrac{\left(2^2\cdot3\right)^{15}\cdot3^4-\left(2^2\right)^5\cdot3^9}{\left(3^3\right)^3\cdot2^{10}-\left(2^5\right)^3\cdot3^9}\\ =\dfrac{2^{30}\cdot3^{15}\cdot3^4-2^{10}\cdot3^9}{3^9\cdot2^{10}-2^{15}\cdot3^9}\\ =\dfrac{3^9\cdot2^{10}\left(2^{20}\cdot3^{10}\right)}{3^9\cdot2^{10}\left(1-2^5\right)}\\ =\dfrac{\left(2^2\right)^{10}\cdot3^{10}}{1-32}\\ =\dfrac{\left(2^2\cdot3\right)^{10}}{-31}\\ =\dfrac{-12^{10}}{31}\)

\(B=\dfrac{3}{1^2\cdot2^2}+\dfrac{5}{2^2\cdot3^2}+...+\dfrac{99}{49^2\cdot50^2}\\ =\dfrac{2^2-1^2}{1^2\cdot2^2}+\dfrac{3^2-2^2}{2^2\cdot3^2}+...+\dfrac{50^2-49^2}{49^2\cdot50^2}\\ =\dfrac{1}{1^2}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+...+\dfrac{1}{49^2}-\dfrac{1}{50^2}\\ =1-\dfrac{1}{2500}\\ =\dfrac{2499}{2500}\)

a) \(\dfrac{\left(-3\right)^7\cdot2^8}{6^7}\)

\(=\dfrac{-1\cdot3^7\cdot2^8}{\left(2\cdot3\right)^7}=\dfrac{-1\cdot3^7\cdot2^7\cdot2}{2^7\cdot3^7}=-1\cdot2=-2\)

b) \(\dfrac{-3\cdot7^4+7^3}{7^5\cdot6-7^3\cdot2}\)

\(=\dfrac{-3\cdot7\cdot7^3+7^3}{7^3\cdot7^2\cdot6-7^3\cdot2}\)

\(=\dfrac{7^3\left(-3\cdot7+1\right)}{7^3\left(7^2\cdot6-2\right)}=\dfrac{-3\cdot7+1}{7^2\cdot6-2}\)

\(=\dfrac{-21+1}{294-2}=\dfrac{-20}{290}=\dfrac{-2}{29}\)

b) \(\dfrac{5^3\cdot3^5}{5^3\cdot0,5+125\cdot2\cdot5}\)

\(=\dfrac{5^3\cdot3^5}{5^3\cdot0,5+5^3\cdot2\cdot5}=\dfrac{5^3\cdot3^5}{5^3\left(0,5+2\cdot5\right)}\)

\(=\dfrac{3^5}{0,5+2\cdot5}=\dfrac{243}{10,5}=\dfrac{162}{7}\)

a)\(\dfrac{6^6+6^3.3^3+3^6}{-73}\)\(=\dfrac{2^6.3^6+2^3.3^3.3^3+3^6}{-73}=\dfrac{2^6.3^6+2^3.3^6+3^6.1}{-73}\)

\(=\dfrac{3^6.\left(2^6+2^3+1\right)}{-73}=\dfrac{3^6\left(64+8+1\right)}{-73}=^{ }\)\(\dfrac{3^6.73}{-73}=\dfrac{3^6}{-1}=-3^6\)

b)\(\dfrac{8^{20}+4^{20}}{4^{25}+64^5}=\dfrac{2^{60}+2^{40}}{2^{50}+2^{30}}=\dfrac{2^{40}.\left(2^{20}+1\right)}{2^{30}.\left(2^{20}+1\right)}=\dfrac{2^{40}}{2^{30}}=2^{10}=1024\)

c)\(\dfrac{45^{10}.5^{20}}{75^{15}}=\dfrac{\left(3^2\right)^{10}.5^{10}.5^{20}}{\left(5^2\right)^{15}.3^{15}}=\dfrac{3^{20}.5^{30}}{5^{30}.3^{15}}=3^5=243\)

a: \(A=\dfrac{2^{12}\cdot3^{10}+2^3\cdot2^9\cdot3^9\cdot3\cdot5}{2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}\)

\(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{11}\cdot3^{11}\cdot7}\)

\(=\dfrac{2^{12}\cdot3^{10}\cdot6}{2^{11}\cdot3^{11}\cdot7}=\dfrac{2}{3}\cdot\dfrac{6}{7}=\dfrac{12}{21}=\dfrac{4}{7}\)

b: \(B=\left(\dfrac{12}{105}+\dfrac{9^{15}}{3}\right)\cdot\dfrac{1}{3}\cdot\dfrac{6^8}{6^4\cdot2^4}\)

\(=\dfrac{12+35\cdot9^{15}}{105}\cdot\dfrac{1}{3}\cdot3^4\)

\(=\dfrac{12+35\cdot9^{15}}{105}\cdot3^3=\dfrac{9\left(12+35\cdot9^{15}\right)}{35}\)

M=\(\dfrac{8^{20}+4^{20}}{4^{25}+64^5}=\dfrac{4^{20}\left(2^{20}+1\right)}{4^{25}+4^{15}}=\dfrac{4^{20}\left(2^{20}+1\right)}{4^{15}\left(4^{10}+1\right)}=\dfrac{2^{20}+1}{4^{10}+1}\)

T=\(\dfrac{45^{10}.5^{20}}{75^{15}}=\dfrac{9^{10}.5^{30}}{25^{15}.3^{15}}=\dfrac{3^{20}.5^{30}}{5^{30}.3^{15}}=3^5=243\)

-3^7.2^8/2^.3^7

=-3.2

=-6

5^3.3^5/5^3(0,5+2,5)

=5^3.3^5/5^3.3\

3^4

=81

5.7^4+7^3.25/7^5.125-7^3.50

=5.7^3(7+5

5.7^4+7^3.25/7^5.125-7^3.50

=5.7^4+7^3.5^2/7^5.5^3-7^3.11.5

=5.7^3(1.7+1.5)/7^3.5(7^2.25-11)

12/1250

emm bấm máy tính câu cuôi

\(\left(\dfrac{2}{3}\right)^3.\left(\dfrac{-3}{4}\right)^2.\left(-1\right)^{2013}=\dfrac{8}{27}.\dfrac{9}{16}.\left(-1\right)=-\dfrac{1}{6}\)

\(\left(\dfrac{1}{5}\right)^{15}.\left(\dfrac{1}{4}\right)^{20}=\dfrac{1}{5^{12}}.\dfrac{1}{4^{20}}=5^{-12}.4^{-20}=125^{-4}.1024^{-4}=\left(125.1024\right)^{-4}=128000^{-4}\)

\(\dfrac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}=\dfrac{2^{12}.3^{10}+2^3.3.5.2^9.3^9}{2^{12}.3^{12}+2^{11}.3^{11}}=\dfrac{2^{12}.3^{10}+2^{12}.2^{10}.5}{2^{12}.3^{12}+2^{11}.3^{11}}=\dfrac{2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^{11}\left(2.3+1\right)}=\dfrac{2.6}{3.7}=\dfrac{4}{7}\)

1: \(=5^{20}\cdot\left(\dfrac{1}{5}\right)^{20}+\left(\dfrac{-3}{4}\cdot\dfrac{-4}{3}\right)^8-1\)

=1+1-1=1

2: \(=\dfrac{15-8}{6}\cdot\dfrac{6}{7}+\left(-\dfrac{3}{2}\right)^2\)

=1+9/4

=13/4

3: \(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{3^8\cdot2^{10}+2^{10}\cdot3^8\cdot5}\)

\(=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{3^8\cdot2^{10}\cdot6}=\dfrac{-2}{6}=\dfrac{-1}{3}\)