Ta có :1990^10 + 1990^9 = 1990^9 . (1990+1) = 1990^9 . 1991 
và 1991^10 = 1991^9 . 1991 
Vì 1990 < 1991 nên 1990^10 + 1990^9 < 1991^10

Thanks 

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 \(2^{x+1}\cdot3^{y}=12^{x}\)

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

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

Câu 1.99²⁰và 9999¹⁰

⁼(99²)¹⁰=9801¹⁰

Vậy 9801¹⁰<9999¹⁰ suy ra  99²⁰<9999¹⁰

Câu 2. 3⁵⁰⁰và 7³⁰⁰

 3⁵⁰⁰=(3⁵)¹⁰⁰=243¹⁰⁰

7³⁰⁰=(7³)¹⁰⁰=343¹⁰⁰

Vậy 243¹⁰⁰<343¹⁰⁰ => 3⁵⁰⁰<7³⁰⁰

\(1990^{10}>1990^9\left(1\right)\)

Ta có  \(1991^1=1990^1+1990^0\)

mà \(\)\(1990^1+1990^0< 1990^9\left(1990>1\right)\)

\(\Rightarrow1990^9>1991^1\left(2\right)\)

\(\left(1\right),\left(2\right)\Rightarrow1991^1< 1990^9< 1990^{10}\)

3⁵=3.3.3.3.3=243

5³=5.5.5=125

Vì: 243> 125=> 3⁵> 5³ (điều phải chứng minh)

3⁵=3.3.3.3.3=243

5³=5.5.5=125

Vì: 243> 125=> 3⁵> 5³ (điều phải chứng minh)