a) \(A=2+2^2+..........+2^{10}\)
\(\Leftrightarrow2A=2^2+2^3+..........+2^{10}+2^{11}\)
\(\Leftrightarrow2A-A=\left(2^2+2^3+.......+2^{11}\right)-\left(2+2^2+.......+2^{10}\right)\)
\(\Leftrightarrow A=2^{11}-2\)
\(\Leftrightarrow A+2=2^{11}\)
\(\Leftrightarrow2^{11}=2^x\)
\(\Leftrightarrow x=11\left(tm\right)\)
Vậy ..............
Câu b tương tự!
1.
a) Ta có \(A=2+2^2+...+2^{10}\)
\(2A=2^2+2^3+2^4+...+2^{10}+2^{11}\)
\(A=2A-A=2^{11}-2\)
thay \(A=2^{11}-2\) vào \(A+2=2^x\) ta có :
\(2^{11}-2+2=2^x\)
=> \(2^{11}=2^x\) \(\Rightarrow x=11\)
Vậy x=11
tik mik nha
\(A=2+2^2+...+2^{10}\)
\(\Rightarrow2A=2\left(2+2^2+...+2^{10}\right)\)
\(\Rightarrow2A=2^2+2^3+...+2^{11}\)
\(\Rightarrow2A-A=\left(2^2+2^3+...+2^{11}\right)-\left(2+2^2+...+2^{10}\right)\)
\(\Rightarrow A=2^{11}-2\)
\(\Rightarrow A+2=2^{11}-2+2\)
\(\Rightarrow2^x=2^{11}\Rightarrow x=11\)
\(D=3+3^2+...+3^{100}\)
\(\Rightarrow3D=3\left(3+3^2+...+3^{100}\right)\)
\(\Rightarrow3D=3^2+3^3+...+3^{101}\)
\(\Rightarrow3D-D=\left(3^2+3^3+...+3^{101}\right)-\left(3+3^2+...+3^{100}\right)\)
\(\Rightarrow2D=3^{101}-3\)
\(\Rightarrow D=\dfrac{3^{101}-3}{2}\)
\(\Rightarrow D+3=\dfrac{3^{101}-3}{2}+3\)
\(\Rightarrow\dfrac{3^{101}-3}{2}+3=3^x+1\)
\(\Rightarrow\dfrac{3^{101}-3}{2}+2=3^x\)
\(\Rightarrow\) đến đoạn này hình như đề sai
