\(=5^2\cdot\dfrac{5}{3}\cdot\dfrac{3}{5}\cdot\dfrac{3}{5}=5^2\cdot\dfrac{3}{5}=5\cdot3=15^1\)

a: \(\left(\dfrac{1}{5}\right)\cdot\left(\dfrac{1}{5}\right)^{15}=\left(\dfrac{1}{5}\right)^{1+15}=\left(\dfrac{1}{5}\right)^{16}\)

b: \(\left(-10,2\right)^{10}:\left(-10,2\right)^3=\left(-10,2\right)^{10-3}=\left(-10,2\right)^7\)

c: \(\left[\left(-\dfrac{7}{9}\right)^7\right]^8=\left(-\dfrac{7}{9}\right)^{7\cdot8}=\left(-\dfrac{7}{9}\right)^{56}\)

a, \(\frac{25.5^3.1}{25.5^2}=\frac{5^2.5^3.1}{5^2.5^2}=\frac{5^5}{5^4}=5\)

b, \(5^2.3^5.\left(\frac{3}{5}\right)^2=5^2.3^5.\frac{9}{25}=5^2.\frac{9}{25}.3^5=9.3^5=3^2.3^5=3^7\)

c, \(\left(\frac{1}{7}\right)^2\cdot\frac{1}{7}\cdot\frac{49}{2}=\frac{1}{49}\cdot\frac{1}{7}\cdot\frac{49}{2}=\frac{1}{49}\cdot\frac{49}{2}\cdot\frac{1}{7}=\frac{1}{2}\cdot\frac{1}{7}=\frac{1}{14}\)

A=5⁰+5¹+...+5⁹⁹

=>5A=5+5²+5³+...+5¹⁰⁰

=>5A-A=4A=(5+5²+...+5¹⁰⁰)-(5⁰+5¹+...+5⁹⁹)=5¹⁰⁰-1

=>4A+1=5¹⁰⁰

Bài 6: 

a: \(2^{27}=8^9\)

\(3^{18}=9^9\)

b: Vì \(8^9< 9^9\)

nên \(2^{27}< 3^{18}\)

\(\begin{array}{l}a){15^8}{.2^4} = {15^{2.4}}{.2^4} = {({15^2})^4}{.2^4}\\ = {225^4}{.2^4} = {(225.2)^4} = {450^4}\\b){27^5}:{32^3} = {({3^3})^5}:{({2^5})^3}\\ = {3^{3.5}}:{2^{5.3}} = {3^{15}}:{2^{15}} = {\left( {\frac{3}{2}} \right)^{15}}\end{array}\)

a) \(15^8\cdot2^4=3^8\cdot5^8\cdot2^4=9^4\cdot25^4\cdot2^4=\left(9\cdot25\cdot2\right)^4=450^4\) 

b) \(27^5:32^3=\left(3^3\right)^5:\left(2^5\right)^3=3^{15}:2^{15}=\left(\dfrac{3}{2}\right)^{15}\)

a) 5³ . 5² . 5 

= 5⁵ . 5 

= 5⁵ . 5¹

= 5⁶

b) 6⁹ : 6⁴ 

= 6⁵

c) 4 . 8 . 16 . 32 

= 2² . 2³ . 2⁴ . 2⁵

= 2⁵ . 2⁴ . 2⁵

= 2⁹ . 2⁵

= 2¹⁴