\(A=8^2\cdot324=8^2\cdot18^2=144^2=\left(12^2\right)^2=12^4\)

\(B=27^3\cdot9^4\cdot243=\left(3^3\right)^3\cdot\left(3^2\right)^4\cdot3^5=3^9\cdot3^8\cdot3^5=3^{22}\)

\(C=6^2\cdot36^4\cdot216^2=6^2\cdot\left(6^2\right)^4\cdot\left(6^3\right)^2=6^2\cdot6^8\cdot6^6=6^{16}\)

\(D=3^4\cdot81^2\cdot9^5=3^4\cdot\left(3^4\right)^2\cdot\left(3^2\right)^5=3^4\cdot3^8\cdot3^{10}=3^{22}\)

GẤP Ạ!!!!!!!!!!!!

\(a,5\cdot p\cdot5\cdot p\cdot2q\cdot4q\)

\(=5^2\cdot p^2\cdot2\cdot q\cdot2^2\cdot q\)

\(=5^2\cdot2^3\cdot p^2\cdot q^2\)

\(b,a\cdot a+b\cdot b+c\cdot c\cdot c+d\cdot d\cdot d\cdot d\)

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

#Urushi

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\(^4\)

ta có :

\(3^3.7^3=\left(3.7\right)^3=21^3\)

3³7³=(3.7)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

a) \(8=2^3\)

\(16=4^2\)

\(27=3^3\)

\(81=9^2\)

\(100=10^2\)

b) \(1000=10^3\)

\(1,000,000=10^6\)

\(1,000,000,000=10^9\)

100.000 } 12 chữ số 0 = 10^12

Bài 6: 

a: \(2^{27}=8^9\)

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

b: Vì \(8^9< 9^9\)

nên \(2^{27}< 3^{18}\)

\(a,23\cdot24\cdot2=2^4\cdot3\cdot23\)

\(b,52\cdot57\cdot53=2^3\cdot3\cdot13\cdot19\cdot53\)

\(c,30\cdot35\cdot37=2\cdot3\cdot5^2\cdot7\cdot37\)

ko vieets dưới dạng lũy thừa nha em

a}23.24.2

=552.2

=1104

b)52.57.53

=2964.53

=157092

c)30.35.37

=1050.37

=38850