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1. Ta có :
\(4A=\frac{2^2\left(2^{18}-3\right)}{2^{20}-3}=\frac{2^{20}-12}{2^{20}-3}=\frac{2^{20}-3-9}{2^{20}-3}=\frac{2^{20}-3}{2^{20}-3}-\frac{9}{2^{20}-3}=1-\frac{9}{2^{20}-3}\)
\(4B=\frac{2^2\left(2^{20}-3\right)}{2^{22}-3}=\frac{2^{22}-12}{2^{22}-3}=\frac{2^{22}-3-9}{2^{22}-3}=\frac{2^{22}-3}{2^{22}-3}-\frac{9}{2^{22}-3}=1-\frac{9}{2^{22}-3}\)
Vì \(2^{20}-3< 2^{22}-3\)
\(\Leftrightarrow\frac{9}{2^{20}-3}>\frac{9}{2^{22}-3}\)
\(\Leftrightarrow1-\frac{9}{2^{20}-3}< 1-\frac{9}{2^{22}-3}\)
\(\Leftrightarrow4A< 4B\)
\(\Leftrightarrow A< B\)
Vậy...
b/ Tương tự
Ta có:
\(A=1+3+3^2+...+3^{2012}\)
\(\Rightarrow3A=3+3^2+3^3+...+3^{2013}\)
\(\Rightarrow3A-A=\left(3+3^2+3^3+...+3^{2013}\right)-\left(1+3+3^2+...+3^{2012}\right)\)
\(\Rightarrow2A=3^{2013}-1\)
\(\Rightarrow A=\left(3^{2013}-1\right):2\)
Do \(B=3^{2013}:2\)
\(\Rightarrow B-A=3^{2013}:2-\left(3^{2013}-1\right):2\)
\(\Rightarrow B-A=\left(3^{2013}-3^{2013}+1\right):2\)
\(\Rightarrow B-A=1:2=\frac{1}{2}\)
Vậy \(B-A=\frac{1}{2}\)