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\(\frac{60^8}{12^7.5^8}=\frac{2^{16}.3^8.5^8}{2^{14}.3^7.5^8}=\frac{2^2.3}{1}=12\)
\(\frac{50^{15}.13^{18}}{25^7.26^{15}.5^{16}}=\frac{2^{15}.5^{30}.13^{18}}{5^{30}.2^{15}.13^{15}}=2197\)
\(\frac{10^4}{5^3}=\frac{5^4.2^4}{5^3}=80\)
\(\frac{60^8}{12^7.5^8}=\frac{2^3.3}{1}=12\)
\(\frac{50^{15}.13^{18}}{25^7.26^{15}}=2197\)
\(10^4:5^3=10000:125=80\)
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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) \(C=\frac{4^{20}+4^{29}}{8^{13}+8^{15}}=\frac{\left(2^2\right)^{20}+\left(2^2\right)^{29}}{\left(2^3\right)^{13}+\left(2^3\right)^{15}}=\frac{2^{40}+2^{58}}{2^{39}+2^{45}}=\frac{2^{40}\left(1+2^{18}\right)}{2^{39}\left(1+2^6\right)}=\frac{2\left(1+2^{18}\right)}{1+2^6}=\frac{2+2^{19}}{1+2^6}\)
\(B=\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{10}}\\ \Rightarrow2B=1+\frac{1}{2}+....=\frac{1}{2^9}\\ \Rightarrow B=1-\frac{1}{2^{10}}\)
\(C=\frac{4^{20}+4^{29}}{8^{13}+8^{15}}=\frac{\left(2^2\right)^{20}+\left(2^2\right)^{29}}{\left(2^3\right)^{13}+\left(2^3\right)^{15}}=\frac{2^{40}+2^{58}}{2^{39}+2^{45}}=\frac{2^{39}\left(2+2^{19}\right)}{2^{39}\left(1+2^6\right)}\)
\(\frac{2^7.16^{13}}{8^{10}.4^{15}}=\frac{2^7.\left(2^4\right)^{13}}{\left(2^3\right)^{10}.\left(2^2\right)^{15}}=\frac{2^7.2^{52}}{2^{30}.2^{30}}=\frac{2^{59}}{2^{60}}=\frac{1}{2}\)