DT
3
K
Khách
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13 tháng 7 2019
\(15.95+5.95-\left(3^2\right)^3+5^7:5^4\)
\(=\left(15+5\right).95-3^6+5^2\)
\(=20.95-729-25\)
\(=1900-729-25\)
\(=1171-25\)
\(=1146\)
15 tháng 7 2019
\(15.95+5.95-\left(3^2\right)^3+5^7:5^4\)
\(=\left(15+5\right)\cdot95-3^6+5^3\)
\(=20\cdot95-729-125\)
\(=1900-729-125\)
\(=1046\)
NN
1 tháng 8 2020
a) \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+........+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.........+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
b) \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+..........+\frac{2}{73.75}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+.......+\frac{1}{73}-\frac{1}{75}\)
\(=\frac{1}{3}-\frac{1}{75}=\frac{8}{25}\)
c) \(\frac{4}{4.6}+\frac{4}{6.8}+\frac{4}{8.10}+..........+\frac{4}{64.66}\)
\(=2.\left(\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+..........+\frac{2}{64.66}\right)\)
\(=2.\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+.....+\frac{1}{64}-\frac{1}{66}\right)\)
\(=2.\left(\frac{1}{4}-\frac{1}{66}\right)=2.\frac{31}{132}=\frac{31}{66}\)
NN
1 tháng 8 2020
d) \(\frac{9}{5.8}+\frac{9}{8.11}+\frac{9}{11.14}+........+\frac{9}{497.500}\)
\(=3.\left(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+..........+\frac{3}{497.500}\right)\)
\(=3.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+......+\frac{1}{497}-\frac{1}{500}\right)\)
\(=3.\left(\frac{1}{5}-\frac{1}{500}\right)=3.\frac{99}{500}=\frac{297}{500}\)
e) \(\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+......+\frac{1}{93.95}\)
\(=\frac{1}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+........+\frac{2}{93.95}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+........+\frac{1}{93}-\frac{1}{95}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{95}\right)=\frac{1}{2}.\frac{18}{95}=\frac{9}{95}\)
g) \(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+..........+\frac{1}{200.203}\)
\(=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+........+\frac{3}{200.203}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+......+\frac{1}{200}-\frac{1}{203}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{203}\right)=\frac{1}{3}.\frac{201}{406}=\frac{67}{406}\)
TN
4
3 tháng 4 2016
nhớ k nha
1/4.7+1/7.10+...+1/73.76=1/3.(3/4.7+3/7.10+..+3/73.76)
=1/3.(1/4-1/7+1/7-1/10+1/10-......+1/73-1/76)
=1/3.(1/4-1/76)
=1/3.9/38=3/38
nhớ k nha
TN
3 tháng 4 2016
\(=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{73}-\frac{1}{76}\)
\(=\frac{1}{4}-\frac{1}{76}\)
\(=\frac{9}{38}\)
TP
2
18 tháng 6 2016
1+2-3-4+5+6-7-8+9+10-11-12+........+298-299-300+301+302 =
1+2+(5-3)+(6-4)+(9-7)+(10-8)+…….+(301-299)+(302-300)=
Từ 302 đến 3 có số cặp là [(302-3):1+1]:2=150 cặp. Mà mỗi cặp có giá trị là 2
Vậy 1+2-3-4+5+6-7-8+9+10-11-12+........+298-299-300+301+302 =
1+2+2×150=3+300=303
NT
1
MH
5 tháng 2 2016
\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)
\(=1-\left(\frac{1}{4}-\frac{1}{4}\right)-\left(\frac{1}{7}-\frac{1}{7}\right)-\left(\frac{1}{10}-\frac{1}{10}\right)-...-\left(\frac{1}{40}-\frac{1}{40}\right)-\frac{1}{43}\)
\(=1-\frac{1}{43}=\frac{42}{43}\)
\(D=\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+...+\frac{1}{73\cdot76}\)
\(D=\frac{1}{3}\left(\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{73\cdot96}\right)\)
\(D=\frac{1}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{73}-\frac{1}{76}\right)\)
\(D=\frac{1}{3}\left(\frac{1}{4}-\frac{1}{76}\right)\)
\(D=\frac{1}{3}\cdot\frac{18}{76}\)
\(=\frac{6}{76}=\frac{3}{38}\)
bằng 3/38
HT