dễ mà : A=2^21-1

cách giải chờ sau đi
 

1) A = 1+2+222 + ... + 22002200

2A = 2 + 222 + 233 + ... + 2201201

2A - A = 2 + 222 +233 + ... + 22012201 - 1 - 2 - ... - 2200200

A = 2201201 - 1

A+1 = 2201201

Vậy a + 1 = 2201201

2) C = 3 + 322 + 333 + ... + 320052005

3C = 322 + 333 + 344 + ... + 320062006

3C - C = 3232 + 333 + 344 + ... + 320062006 - 3 - 322 - 333 - ... - 320052005

2C = 320062006 - 3

2C+3 = 320062006

Vậy 2C + 3 là luỹ thừa của 3 ( Đpcm )

Bài 1:

Ta có: \(A=1+2+2^2+...+2^{200}\)

\(\Leftrightarrow2A=2+2^2+2^3+...+2^{201}\)

\(\Rightarrow2A-A=\left(2+2^2+...+2^{201}\right)-\left(1+2+...+2^{200}\right)\)

\(\Leftrightarrow A=2^{201}-1\)

\(\Rightarrow A+1=2^{201}\)

Bài 2:

Ta có: \(B=3+3^2+3^3+...+3^{2005}\)

\(\Leftrightarrow3B=3^2+3^3+3^4+...+3^{2006}\)

\(\Rightarrow3B-B=\left(3^2+3^3+...+3^{2006}\right)-\left(3+3^2+...+3^{2005}\right)\)

\(\Leftrightarrow2B=3^{2006}-3\)

\(\Rightarrow2B+3=3^{2006}\)

A=2²+2²+2³+2⁴+.........+2²⁰⁰⁵

^{\(2A=2^3+2^3+2^4+2^5+...+2^{2006}\)}

^{\(2A-A=\left(2^3+2^3+2^4+2^5+...+2^{2006}\right)-\left(2^2+2^2+2^3+2^4+...+2^{2005}\right)\)}

^{\(A=\left(2^{2006}+2^3\right)-\left(2^2+2^2\right)\)}

^{\(A=\left(2^{2006}+2^3\right)-2^3\)}

^{\(A=2^{2006}\)}

\(A=2^2+2^2+2^3+2^4+...+2^{2005}\)

\(\Rightarrow2A=2^3+2^3+2^4+2^5+...+2^{2005}+2^{2006}\)

\(\Rightarrow2A-A=\left(2^3+2^3+2^4+2^5+...+2^{2006}\right)-\left(2^2+2^2+2^3+2^4+...+2^{2005}\right)\)

Triệt tiêu hai vế \(\Rightarrow A=\left(2^{2006}+2^3\right)-\left(2^2+2^2\right)=2^{2006}+2^3-2^3\)

\(\Rightarrow A=2^{2006}\)

\(\frac{8}{125}\)không thể viết được dưới dạng một luỹ thừa.

\(\frac{8}{27}=\left(\frac{2}{3}\right)^3\)

a) \(3^5.4^5=\left(3.4\right)^5=12^5\)

b) \(8^5.2^3=\left(2^3\right)^5.2^3=2^{15}.2^3=2^{15+3}=2^{18}\)

\(3^5.4^5=3+4^5=7^5\)

\(8^5.2^3=8+2^{5+3}=10^8\)

Ko biết nữa !

a, 12⁵

b, 2¹⁸

Hok tốt

\(2^6:2^2=2^4=16\)

\(a^5:a=a^4\) chúc bạn học giỏi

ta có 2²=4

4³:4=4^3-1=4²=16

a⁵:a=a^5-1=a⁴

\(^{4^3:2^2=\left(2x2\right)^2:2^2=2^4:2^2=2^2}\)

\(^{a^5xa=a^6}\)