a) 125^80 < 25^118

b) 2^91 > 5^35

c) 107^50 < 73^75

d) 54^4 < 21^12

e) 21^15 < 27^5 . 49^8

k cho mk nha

a)

Ta có :

\(1024^9=\left(2^{10}\right)^9=2^{90}< 2^{100}\)

\(\Rightarrow1024^9< 2^{100}\)

b)

\(\begin{cases}9^{12}=\left(3^2\right)^{12}=3^{24}\\27^7=\left(3^3\right)^7=3^{21}\end{cases}\)

Vì \(3^{24}>3^{21}\)

=> \(9^{12}>27^7\)

c)

\(\begin{cases}125^{80}=\left(5^3\right)^{80}=5^{240}\\25^{118}=\left(5^2\right)^{118}=5^{236}\end{cases}\)

=> 125⁵⁰>25¹¹⁸

c,

9¹² và 27⁷

9¹²= ( 3²)¹² = 3²⁴

27⁷ = ( 3³)⁷ = 3²¹

Vì 3²⁴ > 3²¹ nên 9¹² > 27⁷

^^ Cbht!!!!

\(2^{100}=\left(2^{10}\right)^{10}=1024^{10}>1024^9\)  

=>2^100>1024^9

Bài 1:

a: \(10^{10}=\left(2\cdot5\right)^{10}=2^{10}\cdot5^{10}=2^9\cdot5^{10}\cdot2\)

\(48\cdot50^5=2^4\cdot3\cdot\left(2\cdot5^2\right)^5=2^4\cdot3\cdot2^5\cdot5^{10}=2^9\cdot5^{10}\cdot3\)

mà 2<3

nên \(10^{10}<48\cdot50^5\)

b: \(1990^{10}+1990^9=1990^9\left(1990+1\right)=1990^9\cdot1991\)

\(1991^{10}=1991^9\cdot1991\)

mà 1990<1991

nên \(1990^{10}+1990^9<1991^{10}\)

c: \(107^{50}<108^{50}=\left(2^2\cdot3^3\right)^{50}=2^{100}\cdot3^{150}\)

\(73^{75}>72^{75}=\left(2^3\cdot3^2\right)^{75}=2^{225}\cdot3^{150}\)

mà \(2^{225}\cdot3^{150}>2^{100}\cdot3^{150}=108^{50}>107^{50}\)

nên \(73^{75}>107^{50}\)

d: \(2^{91}=\left(2^{13}\right)^7=8192^7\)

\(5^{35}=\left(5^5\right)^7=3125^7\)

mà 8192>3125

nên \(2^{91}>5^{35}\)

e: \(A=72^{45}-72^{44}=72^{44}\left(72-1\right)=72^{44}\cdot71\)

\(B=72^{44}-72^{43}=72^{43}\left(72-1\right)=72^{43}\cdot71\)

mà 44>43

nên A>B

Bài 2:

a:

ĐKXĐ: x<>2023

\(\frac{x-2023}{4}=\frac{1}{x-2023}\)

=>\(\left(x-2023\right)\left(x-2023\right)=4\cdot1\)

=>\(\left(x-2023\right)^2=4\)

=>\(\left[\begin{array}{l}x-2023=2\\ x-2023=-2\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2+2023=2025\left(nhận\right)\\ x=-2+2023=2021\left(nhận\right)\end{array}\right.\)

b: \(\left(2x+1\right)^4=\left(2x+1\right)^6\)

=>\(\left(2x+1\right)^6-\left(2x+1\right)^4=0\)

=>\(\left(2x+1\right)^4\cdot\left\lbrack\left(2x+1\right)^2-1\right\rbrack=0\)

=>\(\left(2x+1\right)^4\cdot\left(2x+1-1\right)\left(2x+1+1\right)=0\)

=>\(2x\left(2x+1\right)^4\cdot\left(2x+2\right)=0\)

=>\(\left[\begin{array}{l}2x=0\\ 2x+1=0\\ 2x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=-\frac12\\ x=-1\end{array}\right.\)

c: \(\left(3x-1\right)^{10}=\left(3x-1\right)^{20}\)

=>\(\left(3x-1\right)^{20}-\left(3x-1\right)^{10}=0\)

=>\(\left(3x-1\right)^{10}\cdot\left\lbrack\left(3x-1\right)^{10}-1\right\rbrack=0\)

=>\(\left[\begin{array}{l}\left(3x-1\right)^{10}=0\\ \left(3x-1\right)^{10}-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}3x-1=0\\ \left(3x-1\right)^{10}=1\end{array}\right.\)

=>\(\left[\begin{array}{l}3x-1=0\\ 3x-1=1\\ 3x-1=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac13\\ x=\frac23\\ x=0\end{array}\right.\)

d: Sửa đề \(2^{x+1}\cdot3^{y}=12^{x}\)

=>\(2^{x+1}\cdot3^{y}=\left(2^2\cdot3\right)^{x}=2^{2x}\cdot3^{x}\)

=>\(\begin{cases}2x=x+1\\ y=x\end{cases}\Rightarrow\begin{cases}x=1\\ y=x=1\end{cases}\)

a)9^20 và 27^13

9^20=(3^2)^20=3^40

27^13=(3^3)^13=3^39

vì 3^40 > 3^39 =>9^20>27^13

b)10^30 và 2^100

10^30=(10^3)^10=30^10

2^100=(2^10)^10=20^10

vì 30^10>20^0 => 10^30>2^100

c)125^5 và 25^7

125^5=(5^3)^5=5^15

25^7=(5^2)^7=5^14

vì 5^15>5^14 =>125^5>25^7

Ta có :

a) \(9^{20}=\left(3^2\right)^{20}=3^{40};27^{13}=\left(3^3\right)^{13}=3^{39}\)

Vì \(3^{40}>3^{39}\Rightarrow9^{20}>27^{13}\)

Vậy \(9^{20}>27^{13}\)