A=2+2^2+2^3+....+2^10:3

A=(2+2^2)+(2^3+2^4)+....+(2^9+2^10):3

A=2.(1+2)+2^3.(1+2)+...+2^9.(1+2):3

A=2.3+2^3.3+...+2^9.3:3

A=3.(2+2^3+...+2^9):3

vậy A:3 

\(5^{x+1}=125\)

\(5^{x+1}=5^3\)

\(x+1=3\)

\(x=2\)

b, \(5^{2x-3}-2.5^2=5^2.3\)

    \(5^{2x-3}=5^2.\left(3+2\right)\)

    \(5^{2x-3}=5^3\)

    \(2x-3=3\)

     \(x=3\)

c, \(2^x=32\)

    \(2^x=2^5\)

    \(x=5\)

Chúc em học tốt.

a)\(\left(\frac{1}{5}\right)^{10}.5^{20}=\left(\frac{1}{5}\right)^{10}.5^{10.2}=\left(\frac{1}{5}\right)^{10}.25^{10}=\left(\frac{1}{5}.5\right)^{10}=1^{10}=1\)

b)\(5^2.3^5.\left(\frac{3}{5}\right)^2=\left(\frac{3}{5}.5\right)^2.3^5=3^2.3^5=3^7\)

c)\(\left(\frac{1}{16}\right)^3:\left(\frac{1}{8}\right)^2=\left(\frac{1}{8}\right)^{2.3}:\left(\frac{1}{8}\right)^2=\left(\frac{1}{8}\right)^{6+2}=\left(\frac{1}{8}\right)^8\)

\(a.\left(\frac{1}{5}\right)^{10}.5^{20}=\left(\frac{1}{5}\right)^{10}.5^{10.2}=\left(\frac{1}{5}\right)^{10}.\left(5^2\right)^{10}=\left(\frac{1}{5}\right)^{10}.25^{10}=\left(\frac{1}{5}.25\right)^{10}=5^{10}.\)

\(b.5^2.3^5.\left(\frac{3}{5}\right)^2=\left[5^2.\left(\frac{3}{5}\right)^2\right].3^5=\left(5.\frac{3}{5}\right)^2.3^5=3^2.3^5=3^7\)\(c.\left(\frac{1}{16}\right)^3:\left(\frac{1}{8}\right)^2=\left[\left(\frac{1}{4}\right)^2\right]^3:\left[\left(\frac{1}{2}\right)^3\right]^2=\left(\frac{1}{4}\right)^6:\left(\frac{1}{2}\right)^6=\left(\frac{1}{4}:\frac{1}{2}\right)^6=\left(\frac{1}{2}\right)^6\)

bạn tra google là ra ngay ấy mà !!!

hông biết

ta có 

\(4A=4+4^2+4^3+..+4^{99}+4^{100}=\left(1+4+4^2+..+4^{99}\right)+4^{100}-1\)

hay 

\(4A=A+4^{100}-1\Leftrightarrow A=\frac{4^{100}-1}{3}< \frac{4^{100}}{3}=\frac{B}{3}\)

vậy ta có điều phải chứng minh