l) 27^8=(3³)⁸=3²⁴

=>27⁸:3²²=3²⁴:3²²

=3^24-22=3²

m) 8^4=(2³)⁴=2¹²

4⁵=(2²)⁵=2¹⁰

=>8⁴:4⁵=2¹²:2¹⁰

=2^12-10=2²

a) \(27^8:3^{22}\)

\(=\left(3^3\right)^8:3^{22}\)

\(=3^{24}:3^{22}=3^{24-22}\)

\(=3^2\)

b) \(8^4:4^5\)

\(=\left(2^3\right)^4:\left(2^2\right)^5\)

\(=2^{12}:2^{10}=2^{12-10}=2^2\)

a) \(\left(\frac{27}{64}\right)^8:\left(\frac{3}{4}\right)^{22}=\left[\left(\frac{3}{4}\right)^3\right]^8:\left(\frac{3}{4}\right)^{22}=\left(\frac{3}{4}\right)^{24}:\left(\frac{3}{4}\right)^{22}=\left(\frac{3}{4}\right)^2\)

b) \(\left(\frac{2}{3}\right)^2.\left(-\frac{8}{27}\right).\left(-\frac{2}{3}\right)=\left(\frac{2}{3}\right)^2.\left(-\frac{2}{3}\right)^3.\left(-\frac{2}{3}\right)=\left(\frac{2}{3}\right)^2.\left(-\frac{2}{3}\right)^4=\left(\frac{2}{3}\right)^6\)

Ta có:

\(\begin{array}{l}{\left( {\frac{1}{4}} \right)^8} = {[{\left( {\frac{1}{2}} \right)^2}]^8} = {(\frac{1}{2})^{2.8}} = {(\frac{1}{2})^{16}};\\{\left( {\frac{1}{8}} \right)^3} = {[{(\frac{1}{2})^3}]^3} = {(\frac{1}{2})^{3.3}} = {(\frac{1}{2})^9}\end{array}\)

a: \(3\cdot3\cdot3\cdot3\cdot3=3^5\)

b: \(y\cdot y\cdot y\cdot y=y^4\)

c: \(5\cdot p\cdot5\cdot p\cdot2\cdot q\cdot4\cdot q=25\cdot2\cdot4\cdot p^2q^2=2\cdot\left(10qp\right)^2\)

d: \(a\cdot a+b\cdot b+c\cdot c+d\cdot d\cdot d\cdot d=a^2+b^2+c^2+d^4\)

a) 3.3.3.3.3 = 3\(^5\)

b) y.y.y.y = y\(^4\)

c) 5.p.5.p.2.q.4.q = 5\(^2\).p\(^2\).q\(^2\).2\(^3\) 

(2\(^3\) ở đây là vì 2.4 = 2.2.2 = 2\(^3\))

d) a.a + b.b + c.c.c + d.d.d.d = a\(^2\)+b\(^2\)+c\(^3\)+d\(^4\)

g) 3.2.6³ = 3.2.2^3. 3^3 = 2^4 . 3^4

h)4.16.4² = 4 . 4^2 . 4^2 = 4^5

i)3².27.3 = 3^2 * 3^3 *3 = 3^6

k)25⁴:5⁶= 5^8 : 5^6 = 5^2

l)2^1+x=16  

2 * 2^x = 16

2^x = 16:2

2^x = 8

2^x = 2^3

x=3

m)4^x+2=16.4³

4^x . 4^2 = 4^2 . 4^3

4^x = 4^2 . 4^3 : 4^2

4^x = 4^3

x=3

5.p.p.5.p.2.q.q = 5².p³.q².2 

\(5.p.p.5.p^2.q.4.q=\left(5.5.4\right).\left(p.p.p^2\right).\left(q.q\right)=100p^4.q^2\)

5.p.p.5.p².q.4.q

= 5².p⁴.2².q²

= 10².p⁴.q²