a)\(A=1+2^1+2^2+...+2^{2005}\)

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

b) \(2A-A=\left(2+2^2+...+2^{2006}\right)-\left(1+2+2^2+...+2^{2005}\right)\)

\(A=2^{2006}-1\)

             đpcm

a) \(A=1+2+2^2+...+2^{2005}\)

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

b) \(2A-A=\left(2+2^2+2^3+...+2^{2006}\right)-\left(1+2+2^2+...+2^{2005}\right)\)

\(\Rightarrow A=2^{2006}-1\)(đpcm)

_Chúc bạn học tốt_

\(\frac{11}{2}\)+\(x\)=\(\frac{3}{2}\)+\(7\)

\(\frac{11}{2}\)+\(x\)=\(\frac{17}{2}\)

              \(x\)=\(\frac{17}{2}\)-\(\frac{11}{2}\)

               \(x\)=\(\frac{6}{2}\)=3

1\(\frac{1}{2}\)+ X= \(\frac{3}{2}\)- 7.

=> \(\frac{3}{2}\)+ X= \(\frac{-11}{2}\).

=> X= \(\frac{-11}{2}\)- \(\frac{3}{2}\).

=> X=- 7.

Vaayj X=- 7.

\(3^x=9\)

\(\Rightarrow3^x=3^2\)

\(\Rightarrow x=2\)

Cảm ơn nha

\(2^7.3^8.4^9.9^8\)

\(=2^7.3^8.\left(2^2\right)^9.\left(3^2\right)^8\)

\(=2^7.3^8.2^{18}.3^{16}\)

\(=2^{19}.3^{24}\)

\(\left(x^2.y\right)^5.\left(x^2.y^2\right)^7.\left(x.y^2\right)^6.x^3\)

\(=x^{10}.y^5.x^{14}.y^{14}.x^6.y^{12}.x^3\)

\(=x^{33}.y^{31}\)

\(\left|2x-7\right|=x+3\)

\(\Rightarrow2x-7=\hept{\begin{cases}x+3\\-x-3\end{cases}}\)

\(\Rightarrow2x=\hept{\begin{cases}x+10\\-x+4\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=10\\3x=4\Leftrightarrow x=\frac{4}{3}\end{cases}}\)

\(\left(2^3.3^5.5^7\right)^{10}.12^{20}\)

\(=2^{30}.3^{50}.5^{70}.\left(2^2.3\right)^{20}\)

\(=2^{30}.3^{50}.5^{70}.2^{40}.3^{20}\)

\(=2^{70}.3^{70}.5^{70}=\left(30\right)^{70}\)