\(2^{333}< 3^{222}\)

mình cần cách giải

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

a,>

b,>

c,<

a, 19²⁰ > 9⁸

b, 5⁴⁰ < 620¹⁰

c, Ta có: \(2^{161}=2^{7.23}=\left(2^7\right)^{23}=128^{23}\)

=> 128²³ > 13⁴⁰ hay 2¹⁶¹ > 13⁴⁰

1,10²⁰và 90¹⁰

ta có:+,10²⁰=(10²)¹⁰=100¹⁰

        +,90¹⁰=90¹⁰

vì 100¹⁰>90¹⁰=>10²⁰>90¹⁰

2,(1/16)¹⁰ và (1/2)⁵⁰

ta có:+, (1/16)¹⁰=(1/16)¹⁰

         +,(1/2)⁵⁰=(1/2⁵)¹⁰=(1/32)¹⁰

vì (1/16)¹⁰>(1/32)¹⁰=>(1/16)¹⁰>(1/2)⁵⁰

k mik nhé

\(a,\)  \(10^{20}=10^{10+10}=10^{10}.10^{10}\)

        \(90^{10}=9^{10}.10^{10}\)

  Vì \(10^{10}.10^{10}>9^{10}.10^{10}\)

    \(\Rightarrow10^{20}>90^{10}\)

Vậy \(10^{20}>90^{10}\)

\(b,\)\(\left(\frac{1}{16}\right)^{10}=\frac{1^{10}}{16^{10}}=\frac{1}{\left(4^2\right)^{10}}=\frac{1}{4^{20}}\)

   \(\left(\frac{1}{2}\right)^{50}=\frac{1^{50}}{2^{50}}=\frac{1}{\left(2^2\right)^{25}}=\frac{1}{4^{25}}\)

Vì \(\frac{1}{4^{20}}>\frac{1}{4^{25}}\)

\(\Rightarrow\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)

Vậy \(\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)

                         ~~~~~~~~~~Hok tốt~~~~~~~~~~~

10²⁰=100¹⁰

90¹⁰=90¹⁰

Mà 100>90 nên 10²⁰>90¹⁰

Ta thấy : \(10^{20}=100^{10}\)

Với lại \(90^{10}=90^{10}\)

Mà \(100>90=>10^{20}>90^{10}\)

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

nAK.DNX. 0pwi9dOjkciopjopoijasd

=

\(4^{20}=\left(2^2\right)^{20}=2^{40}\)

\(8^{10}=\left(2^3\right)^{10}=2^{30}\)

\(2^{40}>2^{30}\)\(\Rightarrow4^{20}>8^{10}\)

thánh nó trl

\(\text{So sánh : }\)

\(A=2017^{10}+2017^9=1,1149984e33\)

\(B=2018^{10}=1,11998349e33\)

\(\text{Vì : }1,1149984e33< 1,11998349e33\text{ nên }A< B\)