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\(B=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}\)
\(\frac{B}{7}=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.4}\)
\(\frac{B}{7}=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-...+\frac{1}{15}-\frac{1}{28}\)
\(\frac{B}{7}=\frac{1}{2}-\frac{1}{28}\)
\(\frac{B}{7}=\frac{13}{28}\)
\(B=\frac{13}{28}.7\)
\(B=\frac{13}{4}\)
\(n=7\left(\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\right)\)
\(tacocongthuc:\frac{1}{n}-\frac{1}{n+k}=\frac{k}{n\left(n+k\right)}\)
\(\Rightarrow n=7\left(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-.....-\frac{1}{28}\right)=7\left(\frac{1}{2}-\frac{1}{28}\right)=\frac{7.13}{28}=\frac{13}{4}\)
#)Giải :
( k chép lại đề )
\(\frac{n}{7}=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.4}\)
\(\frac{n}{7}=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{15}-\frac{1}{28}\)
\(\frac{n}{7}=\frac{1}{2}-\frac{1}{28}=\frac{13}{28}\)
\(\Rightarrow n=\frac{13}{28}.7=\frac{13}{4}\)
\(\frac{32\cdot2^4}{2^2\cdot\frac{1}{16}}\)
\(=\frac{2^5\cdot2^4}{\frac{4}{16}}\)
\(=\frac{2^9}{\frac{1}{4}}\)
\(=2^9\cdot4\)
\(=2^9\cdot2^2\)
\(=2^{11}\)
\(B=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}\)
\(\frac{B}{7}=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.4}\)
\(\frac{B}{7}=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{15}-\frac{1}{28}\)
\(\frac{B}{7}=\frac{1}{2}-\frac{1}{28}\)
\(\frac{B}{7}=\frac{13}{28}\)
\(B=\frac{13}{28}.7=\frac{13}{4}\)