10³và 2¹⁰⁰

Ta có:10³⁰=(10³)¹⁰=1000¹⁰

          2¹⁰⁰=(2¹⁰)¹⁰=1024¹⁰

Vì 1000<1024 nên 10³⁰<2¹⁰⁰

5³⁰⁰ và 3⁴⁵³

Ta có:5³⁰⁰=(5²)¹⁵⁰=25¹⁵⁰

            3⁴⁵³=(3³)¹⁵¹=27¹⁵¹=27.27¹⁵⁰

Vì  25 < 27.27 nên 5³⁰⁰<3⁴⁵³

nhớ k ch mình nhé

\(2^{27}=2^{3.9}=8^9\)

\(3^{18}=3^{2.9}=9^9\)

Vì \(9^9>8^9\Rightarrow3^{18}>2^{27}\)

MK chỉ làm đc câu a) thui nha :

2^27 = 2^ 3.9 = 8^9

3^18 = 3^2.9=9^9

Vì 9^9 > 8^9 => 2^27 < 2 ^18 

d) 8¹² và 12⁸

Ta có : 8¹² = (9⁶)² = 531441²

           12⁸ = (12⁴)² = 20736²

=> 53441² > 20736²

=> 8¹² >  12⁸

a) 3⁶ và 6³

Ta có : 3⁶ = (3²)³ = 9³

            6³ = 6³

=> 9³ > 6³

=> 3⁶ >  6³

b) ( 0,5)⁶ và ( 0,2)⁹

Ta có : (0,5)⁶ = (0,5)^2*3 = \([\left(0,5\right)^2]^3\)= (0,25)³

          (0,2)⁹ = (0,2)^3*3 = \([\left(0,2\right)^3]^3\)= (0,008)³

    Vì 0,25 > 0,008 => (0,25)³ > (0,008)³

         Vậy (0,5)⁶ > (0,2)⁹

c) (-0,125)⁸ và (0,5)²⁴

Ta có : (-0,125)⁸ = (0,125)⁸

             (0,5)²⁴ = (0,5)^3*8 = \([\left(0,5\right)^3]^8\)= (0,125)⁸

              Vậy (-0,125)⁸ = (0,5)²⁴

a) Ta có \(\hept{\begin{cases}2^{24}=\left(2^6\right)^4=64^4\\3^{16}=\left(3^4\right)^4=81^4\end{cases}}\)

Mà \(64< 81\)

\(\Rightarrow64^4< 81^4\)

\(\Rightarrow2^{24}< 3^{16}\)

b) Ta có \(\hept{\begin{cases}2^{300}=\left(2^3\right)^{100}=8^{100}\\3^{200}=\left(3^2\right)^{100}=9^{100}\end{cases}}\)

Mà 8 < 9  

\(\Rightarrow8^{100}< 9^{100}\)

\(\Rightarrow2^{300}< 3^{200}\)

c) Ta có \(7^{20}=\left(7^4\right)^5=2401^5\)

Ta có 71 < 2401 

\(\Rightarrow71^5< 2401^5\)

\(\Rightarrow71^5< 7^{20}\)

!! K chắc câu c

@@ Học tốt

Chiyuki Fujito

a) \(2^{24}=\left(2^3\right)^8=8^8\)

\(3^{16}=\left(3^2\right)^8=9^8\)

Ta thấy 8<9\(\Rightarrow8^8< 9^8\Rightarrow2^{24}< 3^{16}\)

b) \(2^{300}=\left(2^3\right)^{100}=8^{100}\)

\(3^{200}=\left(3^2\right)^{100}=9^{100}\)

Thấy \(8< 9\Rightarrow8^{100}< 9^{100}\Rightarrow2^{300}< 3^{200}\)

c) \(7^{20}=\left(7^4\right)^5=2401^5\)

Ta thấy \(71< 2401\Rightarrow71^5< 2401^5\Rightarrow71^5< 7^{20}\)

\(\text{a, }2^{30}=8^{10}\)

     \(\text{ }3^{20}=\left(3^2\right)^{10}=9^{10}\)

\(\text{Vậy }2^{30}< 3^{20}\)

\(\text{b, }5^{300}=\left(5^3\right)^{100}=125^{100}\)

     \(3^{500}=\left(3^5\right)^{100}=243^{100}\)

\(\text{Vậy }5^{300}< 243^{100}\)

\(\text{c, }2^{24}=\left(2^3\right)^8=8^8\)

     \(3^{16}=\left(3^2\right)^8=9^8\)

\(\text{Vậy ...}\)

c) Đặt \(A=2^0+2^1+2^2+...+2^{50}\)

\(\Leftrightarrow2A=2^1+2^2+2^3...+2^{51}\)

\(\Leftrightarrow2A-A=2^1+2^2+2^3...+2^{51}\)\(-2^0-2^1-2^2-...-2^{50}\)

\(\Leftrightarrow A=2^{51}-2^0=2^{51}-1< 2^{51}\)

Vậy \(2^0+2^1+2^2+...+2^{50}< 2^{51}\)

a)Ta có: \(\hept{\begin{cases}2^{30}=\left(2^3\right)^{10}=8^{10}\\3^{30}=\left(3^3\right)^{10}=27^{10}\\4^{30}=\left(2^2\right)^{30}=2^{60}\end{cases}}\)và \(\hept{\begin{cases}3^{20}=\left(3^2\right)^{10}=9^{10}\\6^{20}=\left(6^2\right)^{10}=36^{10}\\8^{20}=\left(2^3\right)^{20}=2^{60}\end{cases}}\)

Mà \(8^{10}< 9^{10}\); \(27^{10}< 36^{10}\);\(2^{60}=2^{60}\)nên

\(8^{10}+27^{10}+2^{60}< 9^{10}+36^{10}+2^{60}\)

hay \(2^{30}+3^{30}+4^{30}< 3^{20}+6^{20}+8^{20}\)

b) Ta có: \(4^{30}=2^{30}.2^{30}=8^{10}.4^{15}\)

\(3.24^{10}=3.8^{10}.3^{10}=3^{11}.8^{10}\)

Vì \(4^{15}>3^{11}\) nên \(8^{10}.4^{15}>3^{11}.8^{10}\)

hay \(2^{30}+3^{30}+4^{30}>3.24^{10}\)

a) 3²⁰⁰=(3²)¹⁰⁰=9¹⁰⁰ ; 2³⁰⁰=(2³)¹⁰⁰=8¹⁰⁰

=> 9¹⁰⁰>8¹⁰⁰ hay 3²⁰⁰>2³⁰⁰

b) 71⁵⁰=(71²)²⁵=5041²⁵ ; 37⁷⁵=(37³)²⁵=50653²⁵

=> 5041²⁵<50653²⁵ hay 71⁵⁰<37⁷⁵

c)rút gọn

2016014/2017015=2014/2015

2016016014/2017017015=2014/2015

=> 2014/2015 = 2014/2015

3¹²³⁴ = (3²)⁶¹⁷ = 9⁶¹⁷

2¹⁸⁵¹ = (2³)⁶¹⁷ = 8⁶¹⁷

Ta thấy: 9>8 => 3¹²³⁴ > 2¹⁸⁵¹

nên điền dấu < ban nha

2³⁰+3³⁰+4³⁰>3.24¹⁰