\(A=1+2+2^2+2^3+...+2^{100}\)

\(\Rightarrow2A=2+2^2+2^3+...+2^{200}+2^{201}\)

\(2A-A=A=\left(2+2^2+2^3+...+2^{201}\right)-\left(1+2+2^2+2^3+...+2^{200}\right)\)

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

ta có
A= 1+2+2²+2³+...+2²⁰⁰
2A= 2+2²+2³+...+2²⁰¹
2A-A=(2+2²+2³+...+2²⁰¹)-(1+2+2²+2³+...+2²⁰⁰)
A=2²⁰¹ - 1
A+1=2²⁰¹

 

a, x^7. x^3= x^10

b, (x^2)^5=x^10

c, x^12: x^2= x^10

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

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

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

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

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

\(2^{24}=\left(2^3\right)^{21}=8^{21}\)

2²⁴= 8⁸

`2^24 = 2^(3.8) = (2^3)^8 = 8^8`

A. (-5) ².(-5)⁷

⁼ (-5) ⁽²⁺⁷⁾

= (-5)⁹

B. (0.75)³ :0,75

= (0.75)^(3-1)

= (0.75)²

B. 0,2¹⁰ :0,2⁷

= 0,2^(10-7)

= 0,2³