Lời giải:

$2520.1254=3160080^1$

\(a,4^8.2^{20}=\left(2^2\right)^8.2^{20}=2^{16}.2^{20}=2^{16+20}=2^{36}\\ b,9^{12}.27^5.81^4=\left(3^2\right)^{12}.\left(3^3\right)^5.\left(3^4\right)^4=3^{24}.3^{15}.3^{16}=3^{24+15+16}=3^{55}\\ d,25^{20}.125^4=\left(5^2\right)^{20}.\left(5^3\right)^4=5^{40}.5^{12}=5^{40+12}=5^{52}\\ d,x^7.x^4.x^3=x^{7+4+3}=x^{14}\)

a: 4*8*2^20=2^2*2^3*2^20=2^25

b: 9^12*27^5*81^4=3^24*3^15*3^16=3^55

c: 25^20*125^4=5^40*5^12=5^52

d: =x^(7+4+3)=x^14

giải giúp mình đi minh k cho ma

khong co dap an

358=3.100+5.10+8

     =3.10²+5.10¹+8.10⁰

abcd=a.1000+b.100+c.10+d

       =a.10³+b.10²+c.10¹+d.10⁰

.

987=10 mu 2.9+10 mu1.8+10mu 0.7

2564= 10 mu 3. 2+ 10 mu 2.5 +10 mu 1 .6+ 10 mu 0 . 4

abcd(gach tren dau) = 10mu 3.a+ 10 mu 2 .b + 10 mu 1 .c+10 mu 0.d

xin loi ban nha, minh khong viet duoc luy thua

987=9.100+8.10+7.1                                2564=2.1000+5.100+6.10+4.1           abcde=a.10000+b.1000+c.100+d.10+e.1

     =9.10²+8.10¹+7.10⁰                                    =2.10³+5.10²+6.10¹+4.10⁰ⁿ                   =a.10⁴+b.10³+c.10²+d.10¹+e.10⁰

Phần a = 5mũ10

Phần b  = 9mũ6

a) \(5^6\times5^4=5^{6+4}=5^{10}\)

b)\(9^7\div9^3=9^{7-3}=9^4\)

Đây là số thập phân mà

nhầm là 2^8 nhé

ko ghi lại đề

\(=5^2.\left(5^3\right)^3.\left(5^4\right)^6\)

\(=5^2.5^9.5^{24}\)

\(=5^{35}\)

giúp mình câu b với

koo ghi lại đề

\(=\left(2^5\right)^n.\left(2^6\right)^3.\left(2^7\right)^3\)

\(=2^{5.n}.2^{18}.2^{21}\)

\(=2^{39+5.n}\)

2^8 . 4^5 . 9^9

=2^8 . (2^2)^5 . (3^2)^9

=2^8 . 2^10 . 3^18

=2^18 . 3^18

=(3.2)^18

=6^18

\(2^8.4^5.9^9=2^8.\left(2^2\right)^5.\left(3^2\right)^9\)\(=2^8.2^{10}.3^{18}\)\(=2^{18}.3^{18}\)

\(2^8.4^5.9^9\)

\(=2^8.\left(2^2\right)^5.\left(3^2\right)^9\)

\(=2^8.2^{10}.3^{18}\)

\(=2^{18}.3^{18}\)

\(=\left(2.3\right)^{18}=6^{18}\)