Ta có: 

A = \(\dfrac{10^7+5}{10^7-8}=\dfrac{10^7-8+13}{10^7-8}=1+\dfrac{13}{10^7-8}\)

\(B=\dfrac{10^8+6}{10^8-7}=\dfrac{10^8-7+13}{10^8-7}=1+\dfrac{13}{10^8-7}\)

Mà \(10^8-7>10^7-8\)

=> \(1+\dfrac{13}{10^7-8}>1+\dfrac{13}{10^8-7}\)

=> A < B 

Vậy A < B

Xin lỗi mình kết luận sai vì nhìn nhầm. Đáp án đúng là A > B và cả quá trình trên vẫn đúng nha.

A > B

THAM KHẢO :

A = 53.107 + 54 53 + 1 .107 − 53

= 53.107 + 54 53.107 + 107 − 53

= 53.107 + 54 53.107 + 54 = 1.

B = 134.269 + 135 134 + 1 .269 − 133

= 134.269 + 135 134.269 + 269 − 133

= 134.269 + 135 134.269 + 136 > 1

⇒A < B 

A=B

\(\text{b, 5^36 = (5^3)^12 = 125}^{12}\)

   \(\text{ 11^24 = (11^2)^12}=121^{12}\)

\(\text{Vì }125^{12}>121^{12}=>5^{36}>11^{24}\)

\(\text{c, }107^{50}=\left(107^2\right)^{25}=11449^{25}\)

        \(73^{75}=\left(73^3\right)^{25}=389017^{25}\)

\(\text{Vì }11449^{25}< 389017^{25}\)\(=>107^{50}< 73^{75}\)

D) Ta có : 4³⁶ = (2²)³⁶ = 2^2.36 = 2⁷²

Vì 2⁹¹ > 2⁷² 

=> 2⁹¹ > 4³⁶

119 ngũ 72 lớn

73ngu 75 bé

a) 

Ta có 107⁵⁰ = 107^2x25 = (107²)²⁵ = 11449²⁵

73⁷⁵ = 73^3x25 = (73³)²⁵ = 389017²⁵

vì 11449 < 389017 nên 11449²⁵ < 389017²⁵

Do đó 107⁵⁰ < 73⁷⁵

b) 

Ta có 2⁹¹=(2¹³)⁷=8192⁷

5³⁵=(5⁵)⁷=3125⁷

Vì 8192⁷>3125⁷

Do đó 2⁹¹>5³⁵

c)

Ta có 54⁴ = (2.27)⁴ = (2.3³)⁴ = 2⁴.3¹²
21¹² = (7.3)¹² = 7¹².3¹² 
Vì 7¹² > 2¹² > 2⁴ -

Do đó 54⁴ < 21¹²

a) 107⁵⁰ và 73⁷⁵

Ta có: 107⁵⁰ < 125⁵⁰ và 125⁵⁰ = (5³)⁵⁰= 5¹⁵⁰ 

73⁷⁵ >25⁷⁵ và 25⁷⁵= (5²)⁷⁵= 5¹⁵⁰

Ta có : 107⁵⁰< 125⁵⁰=25⁷⁵<73⁷⁵ suy ra 107⁵⁰< 73⁷⁵

Vậy 107⁵⁰< 73⁷⁵

b)  2⁹¹ và 5³⁵

Ta có: 2⁹¹ > 2⁹⁰ = (2⁵)¹⁸= 32¹⁸

5³⁵<5³⁶ = (5²)¹⁸ = 25¹⁸

Vì 32>25 và 18>0 nên 

2⁹⁰>5³⁶>5³⁵. Mà 2⁹¹> 2⁹⁰ nên 2⁹¹> 5³⁵

c) Ta có: 54⁴ < 60⁴ = 15⁴.4⁴

21¹²>15¹²=15⁴.15⁸

Vì 16⁴.4⁴<15⁴.15⁸ nên 60⁴< 15¹² < 21^{12. Mà} 60⁴ > 56 ⁴ nên 56⁴ < 21¹²

Vập, 56⁴ < 21¹²

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 \(107^{50}=107^{25.2}=\left(107^2\right)^{25}=11449^{25}\)

\(73^{75}=73^{3.25}=\left(73^3\right)^{25}=389017^{25}\)

Vì 11449<389017 => \(11449^{25}

a) Ta có: 107⁵⁰ = 107^2x25 = (107²)²⁵ = 11449²⁵

73⁷⁵ = (73³)²⁵ = 389017²⁵

Mà 11449 < 389017 nên 11449²⁵ < 389017²⁵

=> 107⁵⁰ < 73⁷⁵

b) Ta có: 2⁹¹=(2¹³)⁷= 8192⁷

5³⁵ = (5⁵)⁷ = 3125⁷

Mà 8192⁷>3125⁷

=> 2⁹¹>5³⁵

c) Ta có:  54⁴ = (2.27)⁴ = (2.3³)⁴ = 2⁴.3¹²

21¹² = (7.3)¹² = 7¹².3¹² 

Mà 7¹² > 2¹² > 2⁴ 

=> 54⁴ < 21¹²

b) 2⁹¹ và 5³⁵

2⁹¹ = (2¹³)⁷ = 8192⁷

5³⁵ = ( 5⁵)⁷ = 8125⁷

Vì 8192 > 8125

nên 2⁹¹ > 5³⁵

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

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

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