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

4^30=2^30*2^30

=2^30*4^15

3*24^10=3*3^10*8^10=3^11*2^30

mà 4^30>3^11

nên 2^30+3^30+4^30>3*24^10

Ta có: 4^30=2^30.2^30=2^30.4^15

3.24^10=3.(3.2^3)^10=2^30.3^11

Ta thấy: 3^11<3^15<4^15 => 4^15>3^11

Vì 4^15>3^11 nên 2^30.4^15>2^30.3^11

=>2^30+3^30+4^30>3.24^10

4^30=2^30*2^30

=2^30*4^15

3*24^10=3*3^10*8^10=3^11*2^30

mà 4^30>3^11

nên 2^30+3^30+4^30>3*24^10

4^30=2^30*2^30

=2^30*4^15

3*24^10=3*3^10*8^10=3^11*2^30

mà 4^30>3^11

nên 2^30+3^30+4^30>3*24^10

 có;

4^30=2^30.2^30=(2^3)^10.(2^2)^15=8^10.3^15>8^10.3^11

=8^10.3^10.3=3.24^10

Vậy 2^30.3^30.4^30>3.24^10

****

Ta có:

3.24^10=3^11.4^15 

=> 4^30=4^15.4^15 
 4^15>3^11 (vì phần nguyên bé và mũ cũng bé nên ta có:4^15>3^11)
=>3.24^10<<4^30<<<2^30+3^20+4^30

Ta có :

3.24^10 = 3^11.4^15

=> 4^30 = 4^15.4^15

4^15 > 3^11

=> 3.24^10 < 2^30 + 4^30 + 3^30

4^30=2^30*2^30

=2^30*4^15

3*24^10=3*3^10*8^10=3^11*2^30

mà 4^30>3^11

nên 2^30+3^30+4^30>3*24^10

4³⁰=2³⁰*4¹⁵

3*24¹⁰=3*6¹⁰*4¹⁰=3*3¹⁰*2¹⁰*4¹⁰=3¹¹*2¹⁰*2²⁰=3¹¹*2³⁰

ta co 2³⁰=2³⁰ ma 4¹⁵>3¹¹ => 2³⁰+3³⁰+4³⁰>3*24¹⁰

2³⁰ +3³⁰+4³⁰ > 3.24¹⁰

chứng minh giúp mình đi bạn