5 mũ 91 > 11 mũ 59

2⁹¹>5³⁵

2 mũ 91 > 5 mũ 35

ta có

2 mũ 3 = 8

3 mũ 2=9

vì 8<9 nên 2 mũ 3 <3 mũ 2

27 mũ 11 và 81 mũ 8

625 mũ 5 và 125 mũ 7

5 mũ 36 và 11 mũ 24 

5 mũ 23 và 6,5 mũ 22

7.2 mũ 13 và 2 mũ 16

\(d.2^{105}=\left(2^{7^{ }}\right)^{15}=128^{15}\)

   \(5^{45}=\left(5^3\right)^{15}=125^{15}\)

\(\Rightarrow2^{105}>5^{45}\)

\(e.2^{91}=\left(2^7\right)^{13}=128^{13}\)

    \(5^3=125\)

MÀ 128 > 125 nên 128¹³ > 125

\(\Rightarrow\)2⁹¹ > 5³

Câu f. 2³¹ và 3²¹ mk chưa ra nhưng mk nghĩ là sử dụng tính chất bắc cầu nha !!!!!!!!!!!

a) \(2^{105}=2^{7\times15}=128^{15}\)

\(5^{45}=5^{3\times15}=125^{15}\)

Vậy 2^105 > 5^45

b)\(2^{91}=2^{7\times13}=128^{13}>5^3\)

c)\(2^{31}=2^{30}\times2=2^{3\times10}\times2=8^{10}\times2\)

\(3^{21}=3^{20}\times3=3^{2\times10}\times3=9^{10}\times3\)

\(8^{10}< 9^{10};2< 3\Rightarrow8^{10}\times2< 9^{10}\times3\)

Vậy \(2^{31}< 3^{21}\)

2⁹¹ = ( 2¹³ )⁷ = 8192⁷

5³⁵ = ( 5⁵ )⁷ = 3125⁷

Vì 8192 > 3125 => 2⁹¹ > 5³⁵

$2^{91}>2^{90}={32}^{18}$

         $5^{35}<5^{36}={25}^{18}$

        $\Rightarrow 2^{91}>{32}^{18}>{25}^{18}>5^{35}$

Vậy      $2^{91}>5^{35}$ 

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)Ta có:\(5^{36}=\left(5^3\right)^{12}=125^{12}\)

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

                    Vì \(125^{12}>121^{12}\)\(\Rightarrow5^{36}>11^{24}\)

Câu b mk ko rõ đề bài cho lắm