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\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{13\cdot14}+\frac{1}{14\cdot15}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{14}-\frac{1}{15}\)
\(=\frac{1}{2}-\frac{1}{15}\)
\(=\frac{13}{30}\)
Lời giải:
$3S=10.11(12-9)+11.12(13-10)+12.13(14-11)+...+98.99(100-97)+99.100(101-98)$
$=(10.11.12+11.12.13+12.13.14+...+98.99.100+99.100.101)-(9.10.11+10.11.12+...+97.98.99+98.99.100)$
$=99.100.101-9.10.11$
$\Rightarrow S=\frac{99.100.101-9.10.11}{3}=33.100.101-3.10.11$
\(\dfrac{3}{10.11}\) + \(\dfrac{3}{11.12}\) + \(\dfrac{3}{12.13}\) + \(\dfrac{3}{13.14}\) + \(\dfrac{3}{14.15}\)
= \(\dfrac{3}{10}\) - \(\dfrac{3}{11}\) + \(\dfrac{3}{11}\) - \(\dfrac{3}{12}\) + \(\dfrac{3}{12}\) - \(\dfrac{3}{13}\) + \(\dfrac{3}{13}\) - \(\dfrac{3}{14}\) + \(\dfrac{3}{14}\) - \(\dfrac{3}{15}\)
= \(\dfrac{3}{10}\) - \(\dfrac{3}{15}\) = \(\dfrac{1}{10}\)
\(\dfrac{3}{10.11}+\dfrac{3}{11.12}+\dfrac{3}{12.13}+\dfrac{3}{13.14}+\dfrac{3}{14.15}\)
\(=\dfrac{3}{1}.\left(\dfrac{3}{10}-\dfrac{3}{11}+\dfrac{3}{11}-\dfrac{3}{12}+\dfrac{3}{12}-\dfrac{3}{13}+\dfrac{3}{13}-\dfrac{3}{14}+\dfrac{3}{14}-\dfrac{3}{15}\right)\)
\(=\dfrac{3}{1}.\left(\dfrac{3}{10}-\dfrac{3}{15}\right)\)
\(=\dfrac{3}{1}.\left(\dfrac{9}{30}-\dfrac{6}{30}\right)\)
\(=\dfrac{3}{1}.\dfrac{1}{10}\)
\(=\dfrac{3}{10}\)
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 ♥♥!♥♥
\(C=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(C=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(C=7\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(C=7.\frac{3}{35}\)
\(C=\frac{3}{5}\)
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
\(=5\left(\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{14.15}\right)\)
\(5\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-...+\frac{1}{14}-\frac{1}{15}\right)\)
\(5\left(\frac{1}{10}-\frac{1}{15}\right)\)
\(=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\)
|x+3|=|-9|
TH1: x+3=9 => x=9-3 TH2: x+3=-9=> x=-9 -3
x=6 x=-12
A=\(\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{13}-\frac{1}{14}\)=\(\frac{1}{7}-\frac{1}{14}\)=\(\frac{1}{14}\)
B=0
\(\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+\frac{1}{13.14}\)
\(=\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-\frac{1}{14}\)
\(=\frac{1}{7}-\frac{1}{14}=\frac{1}{14}\)
1/11.12 + 1/12.13 + 1/13.14 +...+ 1/43.44
=1/11-1/12+1/12-1/13+1/13-1/14+...+1/43-1/44
=1/11-1/44
=4/44-1/44
=3/44