2/6+2/12+2/20+.............+2/210
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NK
2
T
4 tháng 12 2016
Ta có:\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+..+\frac{2}{210}=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+..+\frac{2}{14.15}\)
\(=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+..+\frac{1}{14}-\frac{1}{15}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{15}\right)\)
\(=1-\frac{2}{15}=\frac{15-2}{15}=\frac{13}{15}\)
4 tháng 12 2016
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0
NT
1
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
18 tháng 7 2023
A = \(\dfrac{1}{2}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\) + \(\dfrac{1}{30}\)+...+\(\dfrac{1}{182}\)+ 210
A = \(\dfrac{1}{1\times2}\) + \(\dfrac{1}{2\times3}\)+\(\dfrac{1}{3\times4}\)+ \(\dfrac{1}{4\times5}\)+...+ \(\dfrac{1}{13\times14}\)+ 210
A = \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\)+ \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\)+ \(\dfrac{1}{3}\)- \(\dfrac{1}{4}\)+ \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\)+...+\(\dfrac{1}{13}\) - \(\dfrac{1}{14}\) + 210
A = 1 - \(\dfrac{1}{14}\) + 210
A = 211 - \(\dfrac{1}{14}\)
A = \(\dfrac{2953}{14}\)
2 tháng 10 2021
b: \(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)
\(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+3^8\cdot2^{10}\cdot5}\)
\(=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^{11}\cdot3^9}\)
\(=\dfrac{1}{2}\cdot\dfrac{-2}{3}=\dfrac{-1}{3}\)
21 tháng 7 2017
+) \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\)
\(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\)
\(\Rightarrow2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\right)\)
\(\Rightarrow A=1-\dfrac{1}{2^{10}}=\dfrac{2^{10}-1}{2^{10}}\)
Vậy \(A=\dfrac{2^{10}-1}{2^{10}}\)
+) \(F=\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}+...+\dfrac{1}{190}\)
\(\Rightarrow\dfrac{1}{2}F=\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+...+\dfrac{1}{380}\)
\(=\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{19.20}=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{19}-\dfrac{1}{20}\)
\(=\dfrac{1}{5}-\dfrac{1}{20}=\dfrac{3}{20}\Rightarrow F=\dfrac{3}{20}:\dfrac{1}{2}=\dfrac{3}{10}\)
Vậy \(F=\dfrac{3}{10}\)
+) \(G=\dfrac{12}{84}+\dfrac{12}{210}+\dfrac{12}{390}+...+\dfrac{12}{2100}\)
\(=\dfrac{4}{28}+\dfrac{4}{70}+\dfrac{4}{130}+...+\dfrac{4}{700}=\dfrac{4}{4.7}+\dfrac{4}{7.10}+...+\dfrac{4}{25.28}\)
\(=\dfrac{4}{3}.\left(\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{25.28}\right)\)
\(=\dfrac{4}{3}.\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{25}-\dfrac{1}{28}\right)\)
\(=\dfrac{4}{3}.\left(\dfrac{1}{4}-\dfrac{1}{28}\right)=\dfrac{4}{3}.\dfrac{3}{14}=\dfrac{2}{7}\)
Vậy \(G=\dfrac{2}{7}\)
21 tháng 7 2017
\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\)
\(2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\)
\(2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\right)\)
\(A=1-\dfrac{1}{2^{10}}=\dfrac{1024-1}{1024}=\dfrac{1023}{1024}\)
\(F=\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}+...+\dfrac{1}{190}\)
\(=\dfrac{2}{30}+\dfrac{2}{42}+\dfrac{2}{56}+...+\dfrac{2}{380}\)
\(=\dfrac{2}{5.6}+\dfrac{2}{6.7}+\dfrac{2}{7.8}+...+\dfrac{2}{19.20}\)
\(=2\left(\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{19.20}\right)\)
\(=2\left(\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{19}-\dfrac{1}{20}\right)\)
\(=2\left(\dfrac{1}{5}-\dfrac{1}{20}\right)=2.\dfrac{3}{20}=\dfrac{3}{10}\)
\(G=\dfrac{12}{84}+\dfrac{12}{210}+\dfrac{12}{390}+...+\dfrac{12}{2100}\)
\(=\dfrac{4}{28}+\dfrac{4}{70}+\dfrac{4}{130}+...+\dfrac{4}{700}\)
\(=\dfrac{4}{4.7}+\dfrac{4}{7.10}+\dfrac{4}{10.13}+...+\dfrac{4}{25.28}\)
\(=\dfrac{4}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{25}-\dfrac{1}{28}\right)\)
\(=\dfrac{4}{3}\left(\dfrac{1}{4}-\dfrac{1}{28}\right)\)
\(=\dfrac{4}{3}.\dfrac{3}{14}=\dfrac{2}{7}\)
TM
1
27 tháng 6 2016
Dãy số cách đều: 2;4;6;8...; b
Có số số hạng là: \(\frac{\left(b-2\right)}{2}+1=\frac{b}{2}\)
Có tổng là: \(\frac{1}{2}\cdot\left(2+b\right)\cdot\frac{b}{2}=210\Leftrightarrow b\left(b+2\right)=840=28\cdot30=28\cdot\left(28+2\right)\)
Do đó, b = 28
KV
27 tháng 7 2023
a) (125 - x) + 25 = 98
125 - x = 98 - 25
125 - x = 73
x = 125 - 73
x = 52
b) 100 : (x + 20) = 68 : 34
100 : (x + 20) = 2
x + 20 = 100 : 2
x + 20 = 50
x = 50 - 20
x = 30
c) (3x + 12) . 5 = 25 . 4 - 10
(3x + 12) . 5 = 100 - 10
(3x + 12) . 5 = 90
3x + 12 = 90 : 5
3x + 12 = 18
3x = 18 - 12
3x = 6
x = 6 : 2
x = 3
d) 210 : (2x - 3) - 20 = 10
210 : (2x - 3) = 10 + 20
210 : (2x - 3) = 30
2x - 3 = 210 : 30
2x - 3 = 70
2x = 70 + 3
2x = 73
x = 73/2
NM
27 tháng 7 2023
a, ( 125 - x ) + 25 = 98
⇒ 125 - x = 73
⇒ x = 52.
Vậy..
b, 100 : ( x + 20 ) = 68 : 34
⇒ 100 : ( x + 20 ) = 2
⇒ x + 20 = 50
⇒ x = 30
Vậy...
c, ( 3 . x + 12 ) . 5 = 25 . 4 - 10
⇒ ( 3 . x + 12 ) . 5 = 90
⇒ 3 .x + 12 = 18
⇒ 3x = 6
⇒ x = 2.
Vậy...
d, 210 : ( 2 .x - 3 ) - 20 = 10
⇒ ( 2 .x - 3 ) - 20 = 21
⇒ 2x - 3 = 41
⇒ 2x = 44
⇒ x = 22.
Vậy..
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