Xét \(A=2^{30}+3^{30}+4^{30}=\left(2^3\right)^{10}+\left(3^3\right)^{10}+\left(2^2\right)^{30}=8^{10}+27^{10}+2^{60}\)

\(B=3^{20}+6^{20}+8^{20}=\left(3^2\right)^{10}+\left(6^2\right)^{10}+\left(2^3\right)^{20}=9^{10}+36^{10}+2^{60}\)

Vì \(8^{10}< 9^{10},27^{10}< 36^{10}\)nên A<B

2³⁰ = 2^3.10= 8¹⁰

3³⁰=3^3.10=27¹⁰

4³⁰=4^3.10=64¹⁰

Vế trái = 8¹⁰ + 27¹⁰ + 64¹⁰

3²⁰=3^2.10=9¹⁰

6²⁰=6^2.10=36¹⁰

8²⁰=8^2.10=64¹⁰

vế phải = 9¹⁰ + 36¹⁰ + 64¹⁰

Vì 64¹⁰=64¹⁰ ; 36¹⁰ > 27¹⁰ ; 9¹⁰ > 8¹⁰

=> vế phải > vế trái

so sánh: 2^30 + 3^30 + 4^30 và 3^20 + 6^20 + 8^20
2^30 = ( 2^3)^10 = 8^ 10
3^30 = (3^3)^10 = 27^10
4^30 = (4^3)^10 = 64^10
3^20 = (3^2)^10 = 9^10
6^20 = (6^2) = 36^10
8^20 = (8^2)^10 = 84^10
vì 9^10 > 8^10
36^10 > 27^10
84^10 > 64^10
=> 2^30 + 3^30 + 4^30 < 3^20 + 6^20 + 8^20

     <

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

        \(3^{30}=\left(3^3\right)^{10}=27^{10}\)

        \(4^{30}=\left(4^3\right)^{10}=64^{10}\)

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

                  \(6^{20}=\left(6^2\right)^{10}=36^{10}\)

                   \(8^{20}=\left(8^2\right)^{10}=64^{10}\)

              \(\Rightarrow2^{30}+3^{30}+4^{30}=8^{10}+27^{10}+64^{10}\)

            \(\Rightarrow3^{20}+6^{20}+8^{20}=9^{10}+36^{10}+64^{10}\)

       Xét        \(8^{10}

Ta có: \(2^{30}+3^{30}+4^{30}=\left(2^3\right)^{10}+\left(3^3\right)^{10}+\left(4^3\right)^{10}=8^{10}+27^{10}+64^{10}\)

\(3^{20}+6^{20}+8^{20}=\left(3^2\right)^{10}+\left(6^2\right)^{10}+\left(8^2\right)^{10}=9^{10}+36^{10}+64^{10}\)

Vì \(8< 9\)\(\Rightarrow8^{10}< 9^{10}\)

mà \(27< 36\)\(\Rightarrow27^{10}< 36^{10}\)

\(\Rightarrow8^{10}+27^{10}< 9^{10}+36^{10}\)

\(\Rightarrow8^{10}+27^{10}+64^{10}< 9^{10}+36^{10}+64^{10}\)

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

so sánh: 2^30 + 3^30 + 4^30 và 3^20 + 6^20 + 8^20
2^30 = ( 2^3)^10 = 8^ 10
3^30 = (3^3)^10 = 27^10
4^30 = (4^3)^10 = 64^10
3^20 = (3^2)^10 = 9^10
6^20 = (6^2) = 36^10
8^20 = (8^2)^10 = 84^10
vì 9^10 > 8^10
36^10 > 27^10
84^10 > 64^10
=> 2^30 + 3^30 + 4^30 < 3^20 + 6^20 + 8^20

a/ 40^20=40^2.10=1600^10

3^30=3^3.10=27^10

vì 1600^10>27^10 nên 40^20>3^30

a) 40^20=(4^2)^10=16^10

30^30=(3^3)^10=27610

Vì 16<27=>16^10<27^10 hay 4^20<3^30

b) mk chịu

c) Đặt A= 1/3+1/3^2+1/3^3+...+1/3^99

=>3A=3( 1/3+1/3^2+1/3^3+...+1/3^99)

=>3A=1+1/3+1/3^2+...+1/3^98

=>3A-A=(1+1/3+1/3^2+...+1/3^98)-(1/3+1/3^2+1/3^3+...+1/3^99)

=>2A=1-1/3^99

=>A=(1-1/3^99)/2

=>A=1/2 - (1/3^99)/2 < 1/2=>a<1/2

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 \(2^{30}+3^{30}+4^{30}>3.24^{10}\)

cứu tui 

Ta có:\(2^{30}=\left(2^3\right)^{10}=8^{10}< 9^{10}=\left(3^2\right)^{10}=3^{20}\)

\(3^{30}=3^{20}.3^{10}< 3^{20}.4^{10}=3^{20}.\left(2^2\right)^{10}=3^{20}.2^{20}=\left(3.2\right)^{20}=6^{20}\)

\(4^{30}=4^{20}.4^{10}=4^{20}.\left(2^2\right)^{10}=4^{20}.2^{20}=\left(4.2\right)^{20}=8^{20}\)

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

Ta có : x=2³⁰ + 3³⁰ + 4³⁰

x=2^3.10 + 3^3.10 + 4^3.10

x=(2³)²⁰ + (3³)¹⁰ + (4³)¹⁰

x=8²⁰ + 9¹⁰ + 64¹⁰

y=3²⁰ + 6²⁰ + 8²⁰

y=3^2.10 + 6^2.10 + 8^2.10

y=(3²)¹⁰ + (6²)¹⁰ + (8²)¹⁰

y=9¹⁰ + 36¹⁰ + 64¹⁰

mà 9¹⁰ > 8²⁰ ;36¹⁰ >9¹⁰ ;64¹⁰ = 64¹⁰

nên x<y