5/10.11+5/11.12+5/12.13+5/13.14+5/14.15
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5/10.11+5/11.12+5/12.13+5/13.14+5/14.15
=1/10-1/11+1/11-1/12+.....+1/14-1/15
=1/10-1/15
=1/30
\(=5\left(\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{14.15}\right)\)
\(5\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-...+\frac{1}{14}-\frac{1}{15}\right)\)
\(5\left(\frac{1}{10}-\frac{1}{15}\right)\)
\(=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\)
\(\dfrac{3}{10.11}\) + \(\dfrac{3}{11.12}\) + \(\dfrac{3}{12.13}\) + \(\dfrac{3}{13.14}\) + \(\dfrac{3}{14.15}\)
= \(\dfrac{3}{10}\) - \(\dfrac{3}{11}\) + \(\dfrac{3}{11}\) - \(\dfrac{3}{12}\) + \(\dfrac{3}{12}\) - \(\dfrac{3}{13}\) + \(\dfrac{3}{13}\) - \(\dfrac{3}{14}\) + \(\dfrac{3}{14}\) - \(\dfrac{3}{15}\)
= \(\dfrac{3}{10}\) - \(\dfrac{3}{15}\) = \(\dfrac{1}{10}\)
\(\dfrac{3}{10.11}+\dfrac{3}{11.12}+\dfrac{3}{12.13}+\dfrac{3}{13.14}+\dfrac{3}{14.15}\)
\(=\dfrac{3}{1}.\left(\dfrac{3}{10}-\dfrac{3}{11}+\dfrac{3}{11}-\dfrac{3}{12}+\dfrac{3}{12}-\dfrac{3}{13}+\dfrac{3}{13}-\dfrac{3}{14}+\dfrac{3}{14}-\dfrac{3}{15}\right)\)
\(=\dfrac{3}{1}.\left(\dfrac{3}{10}-\dfrac{3}{15}\right)\)
\(=\dfrac{3}{1}.\left(\dfrac{9}{30}-\dfrac{6}{30}\right)\)
\(=\dfrac{3}{1}.\dfrac{1}{10}\)
\(=\dfrac{3}{10}\)
Sửa đề :\(\frac{5}{11.12}+\frac{5}{12.13}+\frac{5}{13.14}+\frac{5}{14.15}\)
\(=5\left(\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}\right)\)
\(=5\left(\frac{1}{11}-\frac{1}{15}\right)\)
\(=\frac{5}{11}-\frac{1}{3}\)
\(=\frac{15-11}{33}=\frac{14}{33}\)
\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{13\cdot14}+\frac{1}{14\cdot15}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{14}-\frac{1}{15}\)
\(=\frac{1}{2}-\frac{1}{15}\)
\(=\frac{13}{30}\)
Lời giải:
$3S=10.11(12-9)+11.12(13-10)+12.13(14-11)+...+98.99(100-97)+99.100(101-98)$
$=(10.11.12+11.12.13+12.13.14+...+98.99.100+99.100.101)-(9.10.11+10.11.12+...+97.98.99+98.99.100)$
$=99.100.101-9.10.11$
$\Rightarrow S=\frac{99.100.101-9.10.11}{3}=33.100.101-3.10.11$
Barack mới tự học toán 6 hơn 1 tuần, có cách giải này
= 102 + 10 + 112 + 11 + 122 + 12 + 132 + 13 + 142 + 14 + 152 + 15 +162 +16 +172 +17 + 182 + 18 + 192 +19
= 102 + 112 + 132 + 142 + 152 + 162 + 172 + 182 +192 + 10 + 11 +12 +13 + 14 + 15 + 16 +17 + 18 + 19
Ta có : (93.2727 - 9393.27).(11.12 + 12.13 + 13.14 + 14.15)
= (93.101.27 - 9393.27).(11.12 + 12.13 + 13.14 + 14.15)
= (9393.27 - 9393.27).(11.12 + 12.13 + 13.14 + 14.15)
= 0.(11.12 + 12.13 + 13.14 + 14.15)
= 0
Vậy ...
Chúc các bn học giỏi ! ❤️
Kb và Tk nhaaa ♥♥!♥♥
\(S1=\dfrac{5}{10.11}+\dfrac{5}{11.12}+.............+\dfrac{5}{14.15}\)
\(\Leftrightarrow S1=5\left(\dfrac{1}{10.11}+\dfrac{1}{11.12}+...............+\dfrac{1}{14.15}\right)\)
\(\Leftrightarrow S1=5\left(\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+.............+\dfrac{1}{14}-\dfrac{1}{15}\right)\)
\(\Leftrightarrow S1=5\left(\dfrac{1}{10}-\dfrac{1}{15}\right)\)
\(\Leftrightarrow S1=5.\dfrac{1}{30}=\dfrac{1}{6}\)
\(S2=\dfrac{1}{5.9}+\dfrac{1}{9.13}+\dfrac{1}{13.17}+........+\dfrac{1}{21.25}\)
\(\Leftrightarrow4S_2=\dfrac{4}{5.9}+\dfrac{4}{9.13}+..............+\dfrac{4}{21.25}\)
\(\Leftrightarrow4S_2=\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+............+\dfrac{1}{21}-\dfrac{1}{25}\)
\(\Leftrightarrow4S_2=\dfrac{1}{5}-\dfrac{1}{25}\)
\(\Leftrightarrow4S_2=\dfrac{4}{25}\)
\(\Leftrightarrow S_2=\dfrac{16}{25}\)
cậu kia làm sai rùi
A .1/5 = 1/10.11+1/11.12+1/12.13+...+1/99,100
A . 1/5 = 1/10-1/11+1/11-1/12+...+1/99-1/100
A .1/5 = 1/10-1/100
A.1/5 = 9/199
A = 9/20
k nhé
\(A=\frac{5}{10.11}+\frac{5}{11.12}+\frac{5}{12.13}+.....+\frac{5}{99.100}\)
=\(\frac{5}{10}-\frac{5}{11}+\frac{5}{11}-\frac{5}{12}+\frac{5}{12}-\frac{5}{13}\)
=\(\frac{5}{10}-\frac{5}{100}=\frac{45}{100}\)=\(\frac{9}{20}\)
A= 5/10.11+5/11.12+5/12.13+5/13.14+5/14.15
A= (1/10.11+1/11.12+1/12.13+1/13.14+1/14.15) :5
A= [(1/10-1/11)+(1/11-1/12)+(1/12-1/13)+(1/13-1/14)+(1/14-1/15)] :5
A= (1/10-1/15):5
A= 1/30:5
A= 1/150