3/10.11+3/11.12+3/12.13+3/13.14+3/14.15
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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
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
\(=5\left(\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{14.15}\right)\)
\(5\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-...+\frac{1}{14}-\frac{1}{15}\right)\)
\(5\left(\frac{1}{10}-\frac{1}{15}\right)\)
\(=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\)
\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{13\cdot14}+\frac{1}{14\cdot15}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{14}-\frac{1}{15}\)
\(=\frac{1}{2}-\frac{1}{15}\)
\(=\frac{13}{30}\)
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 ♥♥!♥♥
Sửa đề :\(\frac{5}{11.12}+\frac{5}{12.13}+\frac{5}{13.14}+\frac{5}{14.15}\)
\(=5\left(\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}\right)\)
\(=5\left(\frac{1}{11}-\frac{1}{15}\right)\)
\(=\frac{5}{11}-\frac{1}{3}\)
\(=\frac{15-11}{33}=\frac{14}{33}\)
|x+3|=|-9|
TH1: x+3=9 => x=9-3 TH2: x+3=-9=> x=-9 -3
x=6 x=-12
\(\dfrac{3}{10.11}\) + \(\dfrac{3}{11.12}\) + \(\dfrac{3}{12.13}\) + \(\dfrac{3}{13.14}\) + \(\dfrac{3}{14.15}\)
= \(\dfrac{3}{10}\) - \(\dfrac{3}{11}\) + \(\dfrac{3}{11}\) - \(\dfrac{3}{12}\) + \(\dfrac{3}{12}\) - \(\dfrac{3}{13}\) + \(\dfrac{3}{13}\) - \(\dfrac{3}{14}\) + \(\dfrac{3}{14}\) - \(\dfrac{3}{15}\)
= \(\dfrac{3}{10}\) - \(\dfrac{3}{15}\) = \(\dfrac{1}{10}\)
\(\dfrac{3}{10.11}+\dfrac{3}{11.12}+\dfrac{3}{12.13}+\dfrac{3}{13.14}+\dfrac{3}{14.15}\)
\(=\dfrac{3}{1}.\left(\dfrac{3}{10}-\dfrac{3}{11}+\dfrac{3}{11}-\dfrac{3}{12}+\dfrac{3}{12}-\dfrac{3}{13}+\dfrac{3}{13}-\dfrac{3}{14}+\dfrac{3}{14}-\dfrac{3}{15}\right)\)
\(=\dfrac{3}{1}.\left(\dfrac{3}{10}-\dfrac{3}{15}\right)\)
\(=\dfrac{3}{1}.\left(\dfrac{9}{30}-\dfrac{6}{30}\right)\)
\(=\dfrac{3}{1}.\dfrac{1}{10}\)
\(=\dfrac{3}{10}\)