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\(B=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}\)
\(=\frac{B}{7}=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.4}\)
\(=\frac{B}{7}=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{15}-\frac{1}{28}\)
\(=\frac{B}{7}=\frac{1}{2}-\frac{1}{28}\)
\(=\frac{B}{7}=\frac{13}{28}\)
\(=B=\frac{13}{28}.7\)
\(=B=\frac{13}{4}\)
\(\frac{1}{7}\)B=\(\frac{5}{2.7.1}+\frac{4}{1.7.11}+\frac{3}{11.2.7}+\frac{1}{2.7.15}+\frac{13}{15.4.7}\)
\(\frac{1}{7}\)B=\(\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(\frac{1}{7}B=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\)
\(\frac{1}{7}B=\frac{1}{2}-\frac{1}{28}\)
\(\frac{1}{7}B=\frac{13}{28}\)
B=\(\frac{13}{28}:\frac{1}{7}\)
B=\(\frac{13}{4}\)
1717B=52.7.1+41.7.11+311.2.7+12.7.15+1315.4.752.7.1+41.7.11+311.2.7+12.7.15+1315.4.7
1717B=52.7+47.11+311.14+114.15+1315.2852.7+47.11+311.14+114.15+1315.28
17B=12−17+17−111+111−114+114−115+115−12817B=12−17+17−111+111−114+114−115+115−128
17B=12−12817B=12−128
17B=132817B=1328
B=1328:171328:17
B=134
1717B=52.7.1+41.7.11+311.2.7+12.7.15+1315.4.752.7.1+41.7.11+311.2.7+12.7.15+1315.4.7
1717B=52.7+47.11+311.14+114.15+1315.2852.7+47.11+311.14+114.15+1315.28
17B=12−17+17−111+111−114+114−115+115−12817B=12−17+17−111+111−114+114−115+115−128
17B=12−12817B=12−128
17B=132817B=1328
B=1328:171328:17
B=134
1717B=52.7.1+41.7.11+311.2.7+12.7.15+1315.4.752.7.1+41.7.11+311.2.7+12.7.15+1315.4.7
1717B=52.7+47.11+311.14+114.15+1315.2852.7+47.11+311.14+114.15+1315.28
17B=12−17+17−111+111−114+114−115+115−12817B=12−17+17−111+111−114+114−115+115−128
17B=12−12817B=12−128
17B=132817B=1328
B=1328:171328:17
B=134
1717B=52.7.1+41.7.11+311.2.7+12.7.15+1315.4.752.7.1+41.7.11+311.2.7+12.7.15+1315.4.7
1717B=52.7+47.11+311.14+114.15+1315.2852.7+47.11+311.14+114.15+1315.28
17B=12−17+17−111+111−114+114−115+115−12817B=12−17+17−111+111−114+114−115+115−128
17B=12−12817B=12−128
17B=132817B=1328
B=1328:171328:17
B=134
1717B=52.7.1+41.7.11+311.2.7+12.7.15+1315.4.752.7.1+41.7.11+311.2.7+12.7.15+1315.4.7
1717B=52.7+47.11+311.14+114.15+1315.2852.7+47.11+311.14+114.15+1315.28
17B=12−17+17−111+111−114+114−115+115−12817B=12−17+17−111+111−114+114−115+115−128
17B=12−12817B=12−128
17B=132817B=1328
B=1328:171328:17
B=134
1717B=52.7.1+41.7.11+311.2.7+12.7.15+1315.4.752.7.1+41.7.11+311.2.7+12.7.15+1315.4.7
1717B=52.7+47.11+311.14+114.15+1315.2852.7+47.11+311.14+114.15+1315.28
17B=12−17+17−111+111−114+114−115+115−12817B=12−17+17−111+111−114+114−115+115−128
17B=12−12817B=12−128
17B=132817B=1328
B=1328:171328:17
B=134
1717B=52.7.1+41.7.11+311.2.7+12.7.15+1315.4.752.7.1+41.7.11+311.2.7+12.7.15+1315.4.7
1717B=52.7+47.11+311.14+114.15+1315.2852.7+47.11+311.14+114.15+1315.28
17B=12−17+17−111+111−114+114−115+115−12817B=12−17+17−111+111−114+114−115+115−128
17B=12−12817B=12−128
17B=132817B=1328
B=1328:171328:17
B=134
1717B=52.7.1+41.7.11+311.2.7+12.7.15+1315.4.752.7.1+41.7.11+311.2.7+12.7.15+1315.4.7
1717B=52.7+47.11+311.14+114.15+1315.2852.7+47.11+311.14+114.15+1315.28
17B=12−17+17−111+111−114+114−115+115−12817B=12−17+17−111+111−114+114−115+115−128
17B=12−12817B=12−128
17B=132817B=1328
B=1328:171328:17
B=134
Ta có :
\(B=\frac{5}{2\cdot1}+\frac{4}{1\cdot11}+\frac{3}{11\cdot2}+\frac{1}{2\cdot15}+\frac{13}{15\cdot4}\)
\(=>\frac{B}{7}=\frac{5}{2\cdot7}+\frac{4}{7\cdot11}+\frac{3}{11\cdot14}+\frac{1}{14\cdot15}+\frac{13}{15\cdot28}\)
\(=>\frac{B}{7}=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\)
\(=>\frac{B}{7}=\frac{1}{2}-\frac{1}{28}=\frac{14}{28}-\frac{1}{28}=\frac{13}{28}\)
\(=>B=\frac{13}{28}\cdot7=\frac{13}{4}\)
B = \(\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}\)
B = \(7.\left(\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\right)\)
B = \(7.\left(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\right)\)
B = \(7.\left(\frac{1}{2}-\frac{1}{28}\right)\)
B = \(7.\frac{13}{28}\)
B = \(\frac{13}{4}\)
\(B=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}\)
\(\Rightarrow\frac{B}{7}=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(\Rightarrow\frac{B}{7}=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\)
\(\Rightarrow\frac{B}{7}=\frac{1}{2}-\frac{1}{28}\)
\(\Rightarrow\frac{B}{7}=\frac{14}{28}-\frac{1}{28}\)
\(\Rightarrow\frac{B}{7}=\frac{13}{28}\)
\(\Rightarrow B=\frac{13}{28}.7\)
\(\Rightarrow B=\frac{13}{4}\)
\(B=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}\)
\(B:7=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}\)
\(\frac{B}{7}=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.4}\)
\(\frac{B}{7}=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{15}-\frac{1}{28}\)
\(\frac{B}{7}=\frac{1}{2}-\frac{1}{28} \)
\(\frac{B}{7}=\frac{13}{28}\)
B=\(\frac{13}{28}\cdot7\)
B=\(\frac{13}{4}\left(=3\frac{1}{4}\right)\)
\(B=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}\)
\(\frac{B}{7}=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.4}\)
\(\frac{B}{7}=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{15}-\frac{1}{28}\)
\(\frac{B}{7}=\frac{1}{2}-\frac{1}{28}\)
\(\frac{B}{7}=\frac{13}{28}\)
\(B=\frac{13}{4}\left(=3\frac{1}{4}\right)\)
Trả lời:
\(\frac{5}{2\cdot1}+\frac{4}{1\cdot11}+\frac{3}{11\cdot2}+\frac{1}{2\cdot15}\)\(+\frac{13}{15\cdot4}\)
\(=\frac{5}{2}+\frac{4}{11}+\frac{3}{22}+\frac{1}{30}\)\(+\frac{13}{60}\)
\(=\frac{55+8+3}{22}\)\(+\frac{2+13}{60}\)
\(=\frac{66}{22}\)\(+\frac{1}{4}\)
\(=\frac{13}{4}\)
Hc tốt #
\(B=\dfrac{5}{2.1}+\dfrac{4}{1.11}+\dfrac{3}{11.2}+\dfrac{1}{2.15}+\dfrac{13}{15.4}\)
\(\dfrac{B}{7}=\dfrac{5}{2.7}+\dfrac{4}{7.11}+\dfrac{3}{11.14}+\dfrac{1}{14.15}+\dfrac{13}{15.4}\)
\(\dfrac{B}{7}=\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{7}{11}+...+\dfrac{1}{15}-\dfrac{1}{28}\)
\(\dfrac{B}{7}=\dfrac{1}{2}-\dfrac{1}{28}\)
\(\dfrac{B}{7}=\dfrac{13}{28}\)
\(B=\dfrac{13}{28}.7=\dfrac{13}{4}\)
\(B=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{2}{2.15}+\frac{13}{15.4}\)
\(B=\frac{1}{2}-1+1-\frac{1}{11}+\frac{1}{11}-\frac{1}{2}+\frac{1}{2}-\frac{1}{15}+\frac{1}{15}-\frac{1}{4}\)
\(B=\frac{1}{2}-\frac{1}{4}\)
\(B=\frac{1}{4}\)
\(B=7\left(\frac{5}{2.1}+\frac{4}{1.11}+...+\frac{13}{15.4}\right)\)
\(B=7\left(\frac{5}{2.7}+\frac{4}{7.11}+...+\frac{13}{15.28}\right)\)
\(B=7\left(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{15}-\frac{1}{28}\right)\)
\(B=7\left(\frac{1}{2}-\frac{1}{28}\right)\)
\(B=7\times\frac{13}{28}\)
\(B=\frac{13}{4}\)