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\(ĐặtA=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)\)
\(A=1-\frac{1}{64}=\frac{63}{64}\)
Đặt: \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2048}\)
\(\Rightarrow A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{11}}\)
\(\Rightarrow2A=2\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{11}}\right)\)
\(=1+\frac{1}{2}+\frac{1}{2^2}+.....+\frac{1}{2^{10}}\)
\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{11}}\right)\)
\(\Rightarrow A=1-\frac{1}{2^{11}}=\frac{2^{11}-1}{2^{11}}=\frac{2047}{2048}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\)
\(2A-A=\left(1+...+\frac{1}{1024}\right)-\left(\frac{1}{2}+...+\frac{1}{2048}\right)\)
\(A=1-\frac{1}{2048}=\frac{2047}{2048}\)
gọi biểu thức là A
A=1/2+1/4+1/8+...+1/2048=1/2+1/2^2+1/2^3+...+1/2^10
=>2A=1+1/2+1/2^2+...+1/2^9
=>A=2A-A(bạn đặt cột dọc ra rồi sẽ thấy:1/2-1/2=0;1/2^2-1/2^2=0;...)Ta được kết quả bằng 1+1/2^10
Đặt A =1/2 + 1/4 + 1/8 + ...+ 1/1024 + 1/2048
A= 1/2 + 1/2^2 + 1/2^3+...+ 1/2^10 + 1/2^11
2A= 1 +1/2 + 1/2^2 +...+ 1/2^9 + 1/2^10
2A-A= (1 +1/2 + 1/2^2 +...+ 1/2^9 + 1/2^10) - (1/2 + 1/2^2 + 1/2^3+...+ 1/2^10 + 1/2^11)
A= 1+1/2 + 1/2^2 +...+ 1/2^9 + 1/2^10 - 1/2 - 1/2^2 - 1/2^3 - ...- 1/2^10 - 1/2^11
A= 1- 1/2^11
A= 2047/ 2048
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{2048}\)
\(=1-\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2048}\)
\(=1-\frac{1}{2048}\)
\(=\frac{2047}{2048}\)
\((6-4\frac{1}{2}):0,03+(3\frac{1}{20}-2,65):4+\frac{2}{5}\)
\((6-4,5):0,03+(3,05-2,65):4+0,4\)
\(=1,5:0,03+0,4:4+0,4\)
\(=50+(0,4\cdot2):4\)
\(=50+0,2\)
\(=50,2\)
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