\(\frac{2^3\cdot5^2\cdot11^2\cdot7}{2^3\cdot5^3\cdot7^2\cdot11}\)

\(=\frac{2^3\cdot5^2\cdot11\cdot11\cdot7}{2^3\cdot5^2\cdot5\cdot7\cdot7\cdot11}\)

\(=\frac{11}{5\cdot7}=\frac{11}{35}\)

ta có 2^3*5^2*11^2*(7/2)^3*5^3*7^2*11

=(2^3*(7/2)^3*7^2)*(5^2*5^3)*(11^2*11)

=(2^3*7^3/2^3*7^2)*5^5*11^3

=7^5*5^5*11^3

\(\dfrac{4^9\cdot36+64^4}{100\cdot16^4}=\dfrac{2^{20}\cdot3^2+2^{24}}{2^{18}\cdot5^2}=\dfrac{2^{20}\left(3^2+2^4\right)}{2^{18}\cdot5^2}=4\)

 ko bt

5.3^x=405

3^x=405:5

3^x=81

3^x=3⁴

^{Vậy x=4}

2^x:8=4

2^x=4.8

2^x=32

2^x=2⁵

^{Vậy x=5}

x²⁸=x⁵

x^28-x^5=0

x^5.x^23-x^5.1=0

x^5.(x^23-1)=0

suy ra x^5=0 hoặc x^23-1=0 suy ra x^5=0^5 hoặc x^23=0+1=1 suy ra x=0 hoặc x^23=1^23 suy ra x=0 hoăc x=1

9

(x-2)^4=256

(x-2)^4=4^4

x-2=4

x=4+2=6

(x+1)^3=125

(x+1)^3=5^3

x+1=5

x=5-1=4

a) \(81^{40}=\left(3^4\right)^{40}=3^{160}\)

   \(27^{14}=\left(3^3\right)^{14}=3^{42}\)

Vì \(3^{160}>3^{42}\) => \(81^{40}>27^{14}\)

b) \(5^{64}=5^{4.16}=625^{16}\)

   \(3^{96}=3^{6.16}=729^{16}\)

Vì \(625^{16}< 729^{16}\)=> \(5^{64}< 3^{96}\)

c) \(125^{12}=\left(5^3\right)^{12}=5^{36}\)

   \(25^{10}=\left(5^2\right)^{10}=5^{20}\)

Vì \(5^{36}>5^{20}\)=> \(125^{12}>25^{10}\)

T_i_c_k nha,mơn bạn nhìu ^^

a, 81⁴⁰>27¹⁴

b, 5⁶⁴<3⁹⁶

c, 125¹²<25¹⁰

a. 81⁴⁰  lớn hơn 27¹⁴

b. 5⁶⁴ bé hơn 3⁹⁶

^c. 125¹² lớn hơn 25¹⁰

k cho mk nhá

sao khong ai tra loi het  z troi - .-

3^x*5^x-1=224

3^x*5^x/5=224

15^x=224*5

15^x=1120

=>ko tồn tại x thỏa mãn đề bài vị 15^x luôn có tận cùng bằng 5 (x khác 0 ) hoặc 1 ( x=0) ma 1120 co tận cùng bằng 0

a) \(9^{21}.9^{33}=9^{21+33}=9^{54}\)

b) \(19^{11}.19.19=19^{11+1+1}=19^{13}\)

c) \(25^2.5^2.125=5^4.5^2.5^3=5^{4+2+3}=5^9\)

d) \(t^{2021}.t^2.\left(t^2\right)^2=t^{2021}.t^2.t^4=t^{2021+2+4}=t^{2027}\)

e) \(123^{14}:123^{13}=123^{14-13}=123\)

f) \(64^2:8^3=\left(8^2\right)^2:8^3=8^4:8^3=8^{4-3}=8=2^3\)

g) \(6^{10}:6^3:36=6^{10}:6^3:6^2=6^{10-3-2}=6^5\)

h) \(m^{20}:m^{10}.m^{10}=m^{20-10+10}=m^{20}\)

oke

oke