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A= 1/2+1/2^2+1/2^3+....+1/2^10(1)
=> 2A = 1+1/2+1/2^2+...+1/2^9(2)
Lấy (2) - (1) ta có ;
=> A = 1-1/2^10
Vậy.................
A = 1/2 + 1/2^2 + 1/2^3 + ...+ 1/2^10
2A = 1 + 1/2 + 1/2^2 + ...+ 1/2^9
2A - A = 1 - 1/2^10
A = 1 - 1/2^10
Chúc học giỏi !!!
Ta có:
\(2A=2.\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
\(=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{11}}\)
\(\Rightarrow2A-A=A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
\(=1-\frac{1}{2^{10}}=\frac{2^{10}-1}{2^{10}}=\frac{1023}{1024}\)
a=1/2+1/2^2+.........+1/2^10
2a=1+1/2+1/2^2+........+1/2^9
2a-a=1-2^10
=1023/1024
2A=2*(1/2+1/2^2+....+1/2^10)
2A=1+1/2+1/2^2+....+1/2^9
2A - A =A=1 - 1/2^10
a) \(\frac{2}{3}x\times\frac{1}{2}=\frac{1}{10}\Rightarrow\frac{2}{3}x=\frac{1}{5}\Rightarrow x=\frac{3}{10}\)
\(A=\dfrac{\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}}{\dfrac{2}{2}+\dfrac{2}{2^2}+...+\dfrac{2}{2^{10}}}=\dfrac{\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}}{2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}\right)}=\dfrac{1}{2}\)
A = \(\dfrac{\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+....+\dfrac{1}{2^{10}}}{\dfrac{2}{2}+\dfrac{2}{2^2}+\dfrac{2}{2^3}+...+\dfrac{2}{2^{10}}}\)
= \(\dfrac{\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}}{2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\right)}\)
= \(\dfrac{1}{2}\)