Tính hợp lí A=(1/3-1).(1/6-1).(1/10-1).....(1/780-1)
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BV
15
XO
8 tháng 6 2020
Ta có A = \(\left(1-\frac{1}{3}\right).\left(1-\frac{1}{6}\right).\left(1-\frac{1}{10}\right)...\left(1-\frac{1}{780}\right)\)
= \(\frac{2}{3}.\frac{5}{6}.\frac{9}{10}...\frac{779}{780}=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}...\frac{1558}{1560}=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{38.41}{39.40}=\frac{\left(1.2.3...38\right).\left(4.5.6...41\right)}{\left(2.3.4...39\right).\left(3.4.5...40\right)}\)
= \(\frac{1.41}{39.3}=\frac{41}{117}\)
VT
8
TN
13 tháng 5 2016
(1-1/3) x (1-1/6) x (1-1/10) x (1-1/15)x ...x (1-1/780)
=2/3x 5/6 x 9/10 x...x 779/780
=4/6 x 10/12 x 18/20 x ...x 1558/1560
=4x10 x 18 x...x 1558/6x 12 x 20 x ...x 1560
= 41/39x 3
= 41/11
DN
4
28 tháng 4 2016
\(B=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)......\left(1-\frac{1}{780}\right)\)
\(B=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}......\frac{779}{780}\)\(=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}.....\frac{1558}{1560}\)
\(B=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.....\frac{38.41}{39.40}\)
\(B=\frac{\left(1.2.3.....38\right)\left(4.5.6.....41\right)}{\left(2.3.4.....39\right)\left(3.4.5.....40\right)}=\frac{1.41}{39.3}=\frac{41}{117}\)
26 tháng 8 2021
\(B=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)......\left(1-\frac{1}{780}\right)\)
\(\Rightarrow B=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}......\frac{779}{780}=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}.....\frac{1558}{1560}\)
\(\Rightarrow B=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.....\frac{38.41}{39.40}\)
\(\Rightarrow B=\frac{\left(1.2.3.....38\right)\left(4.5.6.....41\right)}{\left(2.3.4.....39\right)\left(3.4.5.....40\right)}=\frac{1.41}{39.3}=\frac{41}{117}\)
D
22 tháng 4 2015
a)\(\frac{8}{9}.\frac{15}{16}.\frac{24}{25}.....\frac{2499}{2500}\) \(=\frac{2.4}{3^2}.\frac{3.5}{4^2}.\frac{4.6}{5^2}.....\frac{49.51}{50^2}\)\(=\frac{\left(2.3.4.....49\right)\left(4.5.6....51\right)}{\left(3.4.5.....50\right)\left(3.4.5.....50\right)}=\frac{2.51}{50.3}=\frac{1.17}{25}=\frac{17}{25}\)
D
22 tháng 4 2015
b)\(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)......\left(1-\frac{1}{780}\right)\)
\(=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}......\frac{779}{780}\)\(=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}.....\frac{1558}{1560}\)
\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.....\frac{38.41}{39.40}\)
\(=\frac{\left(1.2.3.....38\right)\left(4.5.6.....41\right)}{\left(2.3.4.....39\right)\left(3.4.5.....40\right)}=\frac{1.41}{39.3}=\frac{41}{117}\)
26 tháng 8 2021
\(D=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)......\left(1-\frac{1}{780}\right)\)
\(\Rightarrow D=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}......\frac{779}{780}=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}.....\frac{1558}{1560}\)
\(\Rightarrow D=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.....\frac{38.41}{39.40}\)
\(\Rightarrow D=\frac{\left(1.2.3.....38\right)\left(4.5.6.....41\right)}{\left(2.3.4.....39\right)\left(3.4.5.....40\right)}=\frac{1.41}{39.3}=\frac{41}{117}\)
NC
0
DA
1
TD
23 tháng 4 2018
= 2/3 . 5/6....779/780
= 4/6 . 10/12.....1558/1560
= 1.4 . 2.5 .... 38.41/ 2.3 . 3. 4. .....39.40
= ( 1.2.3....38).(4.5....41)/(2.3.4....39)(3.4...40)
triệt tiêu xong còn 41/39.3= 41/117
= 4/6 . 10/12.....1558/1560
= 1.4 . 2.5 .... 38.41/ 2.3 . 3. 4. .....39.40
= ( 1.2.3....38).(4.5....41)/(2.3.4....39)(3.4...40)
triệt tiêu xong còn 41/39.3= 41/117
\(A=\left(-\frac{2}{3}\right).\left(-\frac{5}{6}\right).\left(-\frac{9}{10}\right)...\left(\frac{-779}{780}\right)=\left(-\frac{4}{6}\right).\left(-\frac{10}{12}\right).\left(-\frac{18}{20}\right)....\left(-\frac{1558}{1560}\right)\)
\(A=\frac{\left(-1.4\right).\left(-2.5\right).\left(-3.6\right)....\left(-38.41\right)}{\left(2.3\right).\left(3.4\right).\left(4.5\right)....\left(39.40\right)}=\frac{\left(1.2.3....38\right).\left(4.5.6...41\right)}{\left(2.3.4...39\right).\left(3.4.5...40\right)}\) ( Vì từ -1 đến -38 có 38 số =>tích của 38 số âm = tích của 38 số dương)
\(A=\frac{\left(1.2.3....38\right).\left(4.5.6...41\right)}{\left(2.3.4...39\right).\left(3.4.5...40\right)}=\frac{1.41}{39.3}=\frac{41}{117}\)