Tính :
1.5 + 5.9 + 9.13 + .................. + 97.101
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DT
0
DT
0
DT
0
HT
11 tháng 7 2016
\(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{97.101}\)
\(=\frac{1}{4}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{97}-\frac{1}{101}\right)\)
\(=\frac{1}{4}.\left(1-\frac{1}{101}\right)\)
\(=\frac{1}{4}.\frac{100}{101}\)
\(=\frac{25}{101}\)
TH
0
NQ
3
7 tháng 8 2016
1/1.5 + 1/5.9 + 1/9.13 + ... + 1/97.101
= 1/4.(4/1.5 + 4/5.9 + 4/9.13 + ... + 4/97.101)
= 1/4.(1 - 1/5 + 1/5 - 1/9 + 1/9 - 1/13 + ... + 1/97 - 1/101)
= 1/4.(1 - 1/101)
= 1/4.100/101
= 25/101
DD
7 tháng 8 2016
\(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+........+\frac{1}{97.101}\)
\(=\frac{1}{4}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+........+\frac{1}{97}-\frac{1}{101}\right)\)
\(=\frac{1}{4}\left(1-\frac{1}{101}\right)\)
\(=\frac{1}{4}.\frac{100}{101}=\frac{25}{101}\)
BD
15 tháng 6 2023
\(a,\dfrac{3}{5}+\dfrac{3}{5\cdot9}+\dfrac{3}{9\cdot13}+....+\dfrac{3}{97\cdot101}\)
\(=\dfrac{3}{4}\cdot\left(\dfrac{4}{5}+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+....+\dfrac{4}{97\cdot101}\right)\)
\(=\dfrac{3}{4}\cdot\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+....+\dfrac{1}{97}-\dfrac{1}{101}\right)\)
\(=\dfrac{3}{4}\cdot\left(1-\dfrac{1}{101}\right)\)
\(=\dfrac{3}{4}\cdot\dfrac{100}{101}\)
\(=\dfrac{75}{101}\)
\(b,\left(1+\dfrac{1}{2}\right)\cdot\left(1+\dfrac{1}{3}\right)\cdot\left(1+\dfrac{1}{4}\right)\cdot....\cdot\left(1+\dfrac{1}{99}\right)\)
\(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot....\cdot\dfrac{100}{99}\)
\(=\dfrac{100}{2}=50\)
LM
Lê Minh Vũ
CTVHS
15 tháng 6 2023
Tính nhanh:
a) \(\dfrac{3}{5}+\dfrac{3}{5.9}+\dfrac{3}{9.13}+...+\dfrac{3}{97.101}\)
= \(\dfrac{3}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{97}-\dfrac{1}{101}\right)\)
= \(\dfrac{3}{4}\left(1-\dfrac{1}{101}\right)\)
= \(\dfrac{3}{4}\times\dfrac{100}{101}\)
= \(\dfrac{75}{101}\)
b) \(\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{4}\right)...\left(\dfrac{1}{98}+1\right)\left(\dfrac{1}{99}+1\right)\)
\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}...\dfrac{99}{98}.\dfrac{100}{99}\)
\(=\dfrac{3.4.5...99.100}{2.3.4...98.99}\)
\(=\dfrac{100}{2}\)
\(=50\)
NQ
1
21 tháng 3 2020
Đặt \(B=\frac{2}{5\cdot9}+\frac{2}{9\cdot13}+\frac{2}{13\cdot17}+....+\frac{2}{97\cdot101}\)
\(\Rightarrow2B=\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+....+\frac{4}{97\cdot101}\)
\(\Leftrightarrow2B=\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+....+\frac{1}{97}-\frac{1}{101}\)
\(\Leftrightarrow2B=\frac{1}{5}-\frac{1}{101}=\frac{96}{505}\)
\(\Leftrightarrow B=\frac{96}{505}:2\)
viết ..... thế thì ai mà biết
12.A= 1.5.12+5.9(13-1)+9.13(17-5)+13.17(21-9)+.....+97.101(105 - 93)
12.A = 1.5.12 + 5.9.13 -1.5.9 + 9.13.17- 5.9.13 +.....+ 97.101.105 -93.97.101
12.A = 1.5.12 -1.5.9 + 97.101.105
A = (1.5.12 -1.5.9 + 97.101.105):12 = 85725