tính (1-(2/42)).(1-(2/56)).(1-(2/72)).....(1-(2/9900))
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NN
2
12 tháng 5 2022
`1/42 + 1/56 + 1/72 + .... + 1/9900`
`= 1/( 6*7) + 1/( 7*8 ) + ..... + 1/( 99*100)`
`= 1/6 - 1/7 + 1/7-1/8+....+1/99-1/100`
`= 1/6 - 1/100`
`= 50/300 - 3/300`
`= 47/300`
MH
12 tháng 5 2022
\(\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+...+\dfrac{1}{9900}\\ =\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+...+\dfrac{1}{99.100}\\ =\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =\dfrac{1}{6}-\dfrac{1}{100}=\dfrac{47}{300}\)
HM
8 tháng 9 2016
em lớp 6 nha
B= 1/2 + 1/6 + 1/12 +1/20 + 1/30 + 1/42 + 1/56 + 1/72
B= 1/1*2 + 1/2*3 + 1/3*4 + 1/4*5 + 1/5*6 + 1/6*7 + 1/7*8 + 1/8*9
B=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9
B=1+0-0-0-0-0-0-0-1/9
B=1-1/9
B=8/9
k em nha
LT
1
20 tháng 6 2019
\(A=\frac{2}{6}\cdot\frac{12}{20}\cdot\frac{30}{42}\cdot\frac{56}{72}\cdot...\cdot\frac{9900}{101}\)
\(A=\frac{1}{3}\cdot\frac{3}{5}\cdot\frac{5}{7}\cdot\frac{7}{9}\cdot...\cdot\frac{9900}{101}\)
\(A=\frac{1\cdot3\cdot5\cdot7\cdot...\cdot9900}{3\cdot5\cdot7\cdot9\cdot...\cdot101}\)
Đến đây dễ rồi
NH
27 tháng 4 2019
\(A=\frac{1}{30}+\frac{1}{42}+...+\frac{1}{210}\)
\(A=\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{14.15}\)
\(A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{14}-\frac{1}{15}\)
\(A=\frac{1}{5}-\frac{1}{15}\)
Tự tính nha :)
NH
27 tháng 4 2019
\(B=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{99.100}\)
\(B=2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(B=2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(B=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(B=2\left(\frac{1}{2}-\frac{1}{100}\right)\)
Tự làm
DN
3
BT
15 tháng 3 2018
a/ \(A=\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\)
=> \(A=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)
=> \(A=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)
=> \(A=\frac{1}{3}-\frac{1}{9}=\frac{2}{9}\)
BT
15 tháng 3 2018
b/ \(B=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+\frac{2}{10.13}+\frac{2}{13.16}\)
=> \(B=\frac{2}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}\right)\)
=> \(B=\frac{2}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}\right)\)
=> \(B=\frac{2}{3}.\left(\frac{1}{1}-\frac{1}{16}\right)=\frac{2}{3}.\frac{15}{16}=\frac{5}{8}\)
4 tháng 4 2018
\(\left(1-\frac{2}{42}\right)\left(1-\frac{2}{56}\right)...\left(1-\frac{2}{2652}\right)\)
= \(\frac{40}{42}.\frac{54}{56}.\frac{70}{72}...\frac{2650}{2652}\)
= \(\frac{5.8}{6.7}.\frac{6.9}{7.8}.\frac{7.10}{8.9}...\frac{50.53}{51.52}\)
= \(\frac{5.6.7...50}{6.7.8...51}.\frac{8.9.10...53}{7.8.9...52}\)
= \(\frac{5}{51}.\frac{53}{7}=\frac{265}{357}\)
13 tháng 3 2017
-1/90-1/72-1/56-1/42-1/30-1/20-1/12-1/6-1/2
= -1/2-1/6-1/12-1/20-1/30-1/42-1/56-1/64-1/72-1/90
= -(1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/64+1/72+1/90)
= -(1/1x2+1/2x3+1/3x4+1/4x5+1/5x6+1/6x7+1/7x8+1/8x9+1/9x10)
= -(1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10)
= -(1-1/10)
= - 9/10