Question

Thực hiện phép tính :

a) \(2^{2010}\)-(\(2^{2009+}\)\(2^{2008}\)+........+\(2^1\)+\(2^0\)

Tìm x :

b)\(7^{x+2}\)+2.\(7^{x-1}\)=345

Answer 1

a, Đặt \(A=2^{2010}+2^{2009}+2^{2008}+...+2^1+2^0\)

\(\Rightarrow2A=2^{2011}+2^{2010}+2^{2009}+...+2^2+2^1\)

\(\Rightarrow2A-A=2^{2011}-2^0\)

\(\Rightarrow A=2^{2011}-1\)

b,\(7^{x+2}+2.7^{x-1}=345\)

\(7^{x-1}.\left(7^3+2\right)=345\)

\(\Rightarrow7^{x-1}.345=345\)

\(\Rightarrow7^{x-1}=345:345=1\)

\(\Rightarrow7^{x-1}=7^0\)

\(\Rightarrow x-1=0\)

\(\Rightarrow x=1\)

Vậy \(x=1\)