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b) \(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)
\(\Rightarrow2B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
\(\Rightarrow2B-B=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
\(\Rightarrow B=1-\frac{1}{2^9}\)
c) \(C=\frac{4^{20}+4^{29}}{8^{13}+8^{15}}=\frac{\left(2^2\right)^{20}+\left(2^2\right)^{29}}{\left(2^3\right)^{13}+\left(2^3\right)^{15}}=\frac{2^{40}+2^{58}}{2^{39}+2^{45}}=\frac{2^{40}\left(1+2^{18}\right)}{2^{39}\left(1+2^6\right)}=\frac{2\left(1+2^{18}\right)}{1+2^6}=\frac{2+2^{19}}{1+2^6}\)
\(B=\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{10}}\\ \Rightarrow2B=1+\frac{1}{2}+....=\frac{1}{2^9}\\ \Rightarrow B=1-\frac{1}{2^{10}}\)
\(C=\frac{4^{20}+4^{29}}{8^{13}+8^{15}}=\frac{\left(2^2\right)^{20}+\left(2^2\right)^{29}}{\left(2^3\right)^{13}+\left(2^3\right)^{15}}=\frac{2^{40}+2^{58}}{2^{39}+2^{45}}=\frac{2^{39}\left(2+2^{19}\right)}{2^{39}\left(1+2^6\right)}\)
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\(3x.\left(x-\frac{2}{3}\right)=0\)
\(\Leftrightarrow3x=0\)hoặc \(x-\frac{2}{3}=0\)
\(3x=0\Rightarrow x=0\)
\(x-\frac{2}{3}=0\Rightarrow x=0+\frac{2}{3}=\frac{2}{3}\)
Vậy..