C=5/2.1+4/1.11+3/11.2+1/2.15+13/15.4
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5/2.1+4/1.11+3/11.2+1/2.15+13/15.4
-> 5/2+4/11+3/22+1/30+13/60
-> 1650/660+240/660+90/660+22/660+143/660
-> 2145/660
-> 13/4
5/2.1+4/1.11+3/11.2+1/12.15+13/15.4
=7.(5/2.7+4/7.11+3/11.14+1/14.15+13/15.28)
=7.(1/2-1/7+1/7-1/11+1/11-1/14+1/14-1/15+1/15-1/28)
=7.(1/2-1/28)=13/14=hỗn số 3 1 phần 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 )
\(\frac{n}{7}=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.4}\)
\(\frac{n}{7}=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{15}-\frac{1}{28}\)
\(\frac{n}{7}=\frac{1}{2}-\frac{1}{28}=\frac{13}{28}\)
\(\Rightarrow n=\frac{13}{28}.7=\frac{13}{4}\)
B = 7(5/2.7 + 4/7.11 + 3/11.14 + 1/14.15 + 13/15.28)
B = 7.(1/2 - 1/7 + 1/7 - 1/11 + 1/11 - 1/14 + 1/14 - 1/15 + 1/15 - 1/28)
B = 7.( 1/2 - 1/28)
B = 7. 13/28
B = 13/4
k mk đi!
mình không biết đâu nhé
mình mới học lớp 4 thôi
đáp số mình học lớp 4
\(\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.12}+\frac{1}{2.15}+\frac{13}{15.4}\)
\(=\left(\frac{4}{1.11}+\frac{3}{11.12}\right)+\left(\frac{1}{2.15}+\frac{13}{15.4}\right)+\frac{5}{2.1}\)
\(=\frac{1}{11}\left(4+\frac{1}{4}\right)+\frac{1}{15}\left(\frac{1}{2}+\frac{13}{4}\right)+\frac{5}{2.1}\)
\(=\frac{1}{11}.\frac{17}{4}+\frac{1}{15}.\frac{17}{4}+\frac{5}{2}\)
\(=\frac{17}{4}\left(\frac{1}{11}+\frac{1}{15}\right)+\frac{5}{2}\)
\(=\frac{17}{4}.\frac{26}{165}+\frac{5}{2}\)
\(=\frac{442}{660}+\frac{5}{2}\)
\(=\frac{221}{330}+\frac{825}{330}\)
\(=\frac{1046}{330}\)
\(=\frac{523}{165}\)
\(C=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}\)
\(\frac{1}{7}C=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(\frac{1}{7}C=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\)
\(\frac{1}{7}C=\frac{1}{2}-\frac{1}{28}\)
\(\frac{1}{7}C=\frac{13}{28}\)
\(C=\frac{13}{28}:\frac{1}{7}\)
\(C=\frac{13}{4}\)
Vậy \(C=\frac{13}{4}\)