Lời giải:

$A-4=2^2+2^3+2^4+...+2^{20}$

$2(A-4)=2^3+2^4+2^5+....+2^{21}$

$\Rightarrow 2(A-4)-(A-4)=2^{21}-2^2$

$\Rightarrow A-4=2^{21}-4$

$\Rightarrow A=2^{21}$

\(A=2^2+2^3+2^4+...+2^{20}\)
\(2A=2^3+2^4+2^5+...+2^{21}\)
\(A=2A-A=2^{21}-2^2\)

\(A=2+2^2+...+2^{20}\)

\(2A=2^2+2^3+...+2^{21}\)

\(A=2^{21}-2\)

\(A=4+2^2+2^3+.......+2^{20}\)

\(A=4+2^2+2^3+.......+2^{20}\)

\(2A=8+2^{3^{ }}+.........+2^{21}\)

\(2A-A=A=2^{21}+2^{20}+......+8-4-2^2-......-2^{20}\)

\(A=2^{21}\)

1.
a) \(3^4\times3^5\times3^6=3^{4+5+6}=3^{15}\)

b) \(5^2\times5^4\times5^5\times25=5^2\times5^4\times5^5\times5^2=5^{2+4+5+2}=5^{13}\)

c) \(10^8\div10^3=10^{8-3}=10^5\)

d) \(a^7\div a^2=a^{7-2}=a^5\)

2.

\(987=900+80+7\\ =9\times100+8\times10+7\\ =9\times10^2+8\times10^1+7\times10^0\)

\(2021=2000+20+1\\ =2\times1000+2\times10+1\times1\\ =2\times10^3+2\times10^1+1\times10^0\)

\(abcde=a\times10000+b\times1000+c\times100+d\times10+e\times1\\ =a\times10^4+b\times10^3+c\times10^2+d\times10^1+e\times10^0\)

Ta có 

A = 2² + 2² + 2³ + 2⁴ + .. + 2²⁰

A = 2 . 2² + 2³ + 2⁴ +.. + 2²⁰

A = 2³ + 2³ + 2⁴ + .. + 2²⁰ 

A = 2⁴ + 2⁴ + ... 2²⁰ 

Làm như vậy cho đến khi A = 2¹⁹ + 2¹⁹ + 2²⁰

A = 2²⁰  + 2²⁰ = 2 . 2²⁰ = 2²¹

= 2²¹ tick đi mik giải cho 

Ta có A = 2² + 2² + 2³ + 2⁴ + ............ + 2²⁰

=> 2A = 2.(2² + 2² + 2³ + 2⁴ + ............ + 2²⁰)

=> 2A = 2³ + 2³ + 2⁴ + 2⁵ + ............ + 2²¹ 

=> 2A - A = 2²¹ + 2³ - 2² - 2²

=> A = 2²¹ (đpcm)

Ta có A = 2² + 2² + 2³ + 2⁴ + ............ + 2²⁰

=> 2A = 2.(2² + 2² + 2³ + 2⁴ + ............ + 2²⁰)

=> 2A = 2³ + 2³ + 2⁴ + 2⁵ + ............ + 2²¹ 

=> 2A - A = 2²¹ + 2³ - 2² - 2²

=> A = 2²¹ (đpcm)