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\(2A=1+\frac{1}{2}+...+\frac{1}{2^{49}}\)

\(2A-A=1-\frac{1}{2^{50}}\)

\(A=1-\frac{1}{2^{50}}\)=> A bé hơn 1

tương tự nha

\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\)

\(2A=2.\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\right)\)

\(2A=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{48}}+\frac{1}{2^{49}}\)

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{48}}+\frac{1}{2^{49}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\right)\)

\(A=1-\frac{1}{2^{50}}< 1\)

\(B=\frac{1}{2}+\left(\frac{1}{2}\right)^2+....+\left(\frac{1}{2}\right)^{99}\)

\(\Rightarrow2B=1+\frac{1}{2}+...+\left(\frac{1}{2^{98}}\right)\)

\(\Rightarrow B=\frac{1}{2}-\frac{1}{2^{99}}>-\frac{1}{2}>A\)

\(\Rightarrow B>A\)

2410 là 24¹⁰

a) \(4:\left(x-1\right)=\left(x-1\right):9\)

\(\frac{4}{x-1}=\frac{x-1}{9}\)

\(\left(x-1\right)^2=36\)

\(\left(x-1\right)^2=6^2\)

\(\Rightarrow x-1=6\)

\(\Rightarrow x=7\)

vậy \(x=7\)

c) \(3\frac{1}{2}:x\frac{1}{2}=5\frac{1}{3}:\frac{1}{2}.1\frac{1}{5}\)

\(\frac{7}{2}:\frac{1}{2}x=\frac{16}{3}:\frac{1}{2}.\frac{6}{5}\)

\(\frac{7}{2}:\frac{1}{2}x=\frac{64}{5}\)

\(\frac{1}{2}x=\frac{7}{2}:\frac{64}{5}\)

\(\frac{1}{2}x=\frac{35}{128}\)

\(x=\frac{35}{128}:\frac{1}{2}\)

\(x=\frac{35}{64}\)

d) \(\left|2x-3\right|=5\)

\(\Rightarrow\orbr{\begin{cases}2x-3=5\\2x-3=-5\end{cases}}\Rightarrow\orbr{\begin{cases}2x=8\\2x=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-1\end{cases}}\)

vậy \(\orbr{\begin{cases}x=4\\x=-1\end{cases}}\)

f) \(\left(2x-\frac{1}{2}\right)^2=\left(1-3x\right)^2\)

\(\Rightarrow2x-\frac{1}{2}=1-3x\)

\(\Rightarrow2x+3x=1+\frac{1}{2}\)

\(\Rightarrow5x=\frac{3}{2}\)

\(\Rightarrow x=\frac{3}{10}\)

\(\left(3-\frac{1}{2}:x\right)^2=14\)

\(\left(3-\frac{1}{2x}\right)^2=14\)

\(\frac{1}{4x^2}-2.\frac{1}{2x}.3+9=14\)

\(\frac{1}{4x^2}-\frac{3}{x}=5\)

\(\left(\frac{1}{4x}-3\right):x=5\)

y1 và y2 lần lượt bằng 8 và 6

còn x1, x2 lần lượt bằng -4 và -10

tick nhóe!

ahihi

\(1-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^{10}}\)

\(=1-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)(1)

Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\)

\(\Rightarrow2A=1+\frac{1}{2}+...+\frac{1}{2^9}\)

\(\Rightarrow2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)

\(\Rightarrow A=1-\frac{1}{2^{10}}\)

Thay A vào (1)

\(\Rightarrow1-\left(1-\frac{1}{2^{10}}\right)\)

\(=1-1+\frac{1}{2^{10}}=\frac{1}{2^{10}}\)

Ta có: 2¹⁰ < 2¹¹

\(\Rightarrow\frac{1}{2^{10}}>\frac{1}{2^{11}}\)(đpcm)