a) \(\frac{9^{10}.27^7}{81^7.3^{15}}\)= \(\frac{\left(3^2\right)^{10}.\left(3^3\right)^7}{\left(3^4\right)^7.3^{15}}\)= \(\frac{3^{2.10}.3^{3.7}}{3^{4.7}.3^{15}}\)=\(\frac{3^{20}.3^{21}}{3^{28}.3^{15}}\)=\(\frac{3^{41}}{3^{43}}\)= \(\frac{3^{41}}{3^{41}.3^2}\)= \(\frac{1}{3^2}=\frac{1}{9}\)

b) \(\frac{8^3.9^5.27^5}{4^5.81^6}\)= \(\frac{\left(2^3\right)^3.\left(3^2\right)^5.\left(3^3\right)^5}{\left(2^2\right)^5.\left(3^4\right)^6}\)= \(\frac{2^{3.3}.3^{2.5}.3^{3.5}}{2^{2.5}.3^{4.6}}\)=\(\frac{2^9.3^{10}.3^{15}}{2^{10}.3^{24}}\)= \(\frac{2^9.3^{25}}{2^{10}.3^{24}}\)=\(\frac{2^9.3^{24}.3}{2^9.2.3^{24}}\)=\(\frac{3}{2}\)

kệ cha nhà bây

a)   378

b)   3

c)    2

d)    2

e)    \(\frac{8748}{1715}\)

Mình thấy bài e) bạn có ghi thiếu ko vậy.81^2 x;: hay là cộng trừ vậy?

a) 25⁵⁰.125⁵
= (5²)⁵⁰.(5³)⁵
= 5¹⁰⁰.5¹⁵
= 5¹⁰⁰⁺¹⁵
= 5¹¹⁵
b) 9²⁵.27⁴.81³
= (3²)²⁵.(3³)⁴.(3⁴)³
= 3⁵⁰.3¹².3¹²
= 3^50+12+12
= 3⁷⁴
d) 12⁷:6⁷
= (12:6)⁷
= 2⁷
 

a/\(378\)

b/\(3\)

a: \(=8^8\left(8^2-8-1\right)=8^8\cdot55⋮5\)

b: \(=7^4\left(7^2+7+1\right)=7^4\cdot57⋮̸11\)

a.\(12^7:6^7=\left(12:6\right)^7=2^7\)

b. \(27^5:81^3=\left(3^3\right)^5:\left(3^4\right)^3=3^{15}:3^{12}=3^3\)

c. \(18^3:9^3=\left(18:9\right)^3=2^3\)

d. \(125^3:25^4=\left(25^3\right)^3:25^4=25^9:25^4=25^5\)

12^7:6^7=(12:6)^7=2^7

27^5:81^3=(3^3)^5:(3^4)^3=3^15:3^12=3^3

18^3:9^3=(18:9)^3=2^3

125^3:25^4=(5^3)^3:(5^2)^4=5^9:5^8=5

a: \(=\dfrac{1}{9}+\dfrac{13}{4}+5+\dfrac{3}{16}+4+\dfrac{1}{3}+\dfrac{14}{5}+\dfrac{1}{2}\)

\(=9+\dfrac{4}{9}+\dfrac{14}{5}+\dfrac{52}{16}+\dfrac{3}{16}+\dfrac{8}{18}\)

\(=9+\dfrac{146}{45}+\dfrac{63}{16}=\dfrac{11651}{720}\)

b: \(=\dfrac{7}{3}+\dfrac{9}{20}+\dfrac{85}{20}+\dfrac{1}{81}+6+\dfrac{8}{27}\)

\(=6+\dfrac{94}{20}+\dfrac{7\cdot27+1+8\cdot3}{81}\)

\(=6+\dfrac{94}{20}+\dfrac{214}{81}=\dfrac{10807}{810}\)