Ta có :C= 2181-729+243.81-27

=2052+19683-27

C=21108

D=\(3^2.9^2.243+18.243.324.243\)

=9.81.243+18.243.324.243

=177147+344373768

=344550915

Ta có : C:D=21108:344550915=0,00006

9*729=3^2*3^6=3^8

=>3^9>9*729

ko lon hon nha ban

Không, vì 3 mũ 9 không lớn hơn 9 nhân 729

3⁹ = 19683

9 . 729 = 6561

Vậy câu khẳng định ban đầu sai.

729=3⁶=9³=27²

729 = 27² = 9³= 3⁶

`729 = 9^3 = 3^6 = 27^2`

a) ta có: 7^10 < 7^14 = (7^2)^7 = 49^7 < 50^7

=> 7^10 < 50^7

b) ta có: 5^30 = (5^3)^10 = 125^10 > 124^10

=> 5^30 > 124^10

c) ta có: 9^21 = (9^3)^7=729^7

phần d thì mk ko bk, xl bn nha

a,<

b,>

c,=

^_^

Nhu lon

a) \(\left(\frac{1}{4}\right)^x:\frac{1}{16}=\frac{1}{4}\)

\(\left(\frac{1}{4}\right)^x=\frac{1}{4}\cdot\frac{1}{16}\)

\(\left(\frac{1}{4}\right)^x=\frac{1}{64}\)

\(\left(\frac{1}{4}\right)^x=\left(\frac{1}{4}\right)^3\)

\(\Rightarrow x=3\)

b) \(81^x:9^x=729\)

\(\left(81:9\right)^x=729\)

\(9^x=9^3\)

\(\Rightarrow x=3\)

a) \(\left(\frac{1}{4}\right)^x:\frac{1}{16}=\frac{1}{4}\)

\(\Rightarrow\left(\frac{1}{4}\right)^x=\frac{1}{4}\times\frac{1}{16}\)

\(\Rightarrow\left(\frac{1}{4}\right)^x=\frac{1}{64}\)

\(\Rightarrow\left(\frac{1}{4}\right)^x=\left(\frac{1}{4}\right)^3\)

\(\Rightarrow x=3\)

Vậy x = 3

b) \(81^x:9^x=729\)

\(\Rightarrow\left(81:9\right)^x=729\)

\(\Rightarrow9^x=9^3\)

\(\Rightarrow x=3\)

Vậy x = 3

_Chúc bạn học tốt_

\(\left(2x+1\right)^3=729\)

\(\Leftrightarrow\)\(\left(2x+1\right)^3=9^3\)

\(\Leftrightarrow\)\(2x+1=9\)

\(\Leftrightarrow\)\(2x=8\)

\(\Leftrightarrow\)\(x=4\)

Vậy...

\(\left(2x+1\right)^3=729\)

\(\Rightarrow2x+1=9\)

\(\Rightarrow2x=8\)

\(\Rightarrow x=4\)

(2x + 1)³ = 729

(2x + 1)³ = 9³

2x + 1     = 9

2x           = 9 - 1

x             = 8 : 2

x             = 4