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DN
2
17 tháng 5 2022
\(A=\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{8}\right)\left(1+\dfrac{1}{15}\right)...\left(1+\dfrac{1}{2400}\right)\)
\(=\dfrac{4}{3}.\dfrac{9}{8}...\dfrac{2401}{2400}\)
\(=\dfrac{2.2}{1.3}.\dfrac{3.3}{2.4}...\dfrac{49.49}{48.50}\)
\(=\dfrac{2.3...49}{2.3...49}.\dfrac{2.3...49}{3.4...50}=1.\dfrac{2.49}{50}=\dfrac{49}{25}\)
O
17 tháng 5 2022
\(A=\left(1+\dfrac{1}{3}\right)\cdot\left(1+\dfrac{1}{8}\right)\cdot\left(1+\dfrac{1}{15}\right)\cdot\cdot\cdot\left(1+\dfrac{1}{2400}\right)\)
\(A=\left(\dfrac{3}{3}+\dfrac{1}{3}\right)\cdot\left(\dfrac{8}{8}+\dfrac{1}{8}\right)\cdot\left(\dfrac{15}{15}+\dfrac{1}{15}\right)\cdot\cdot\cdot\left(\dfrac{2400}{2400}+\dfrac{1}{2400}\right)\)
\(A=\dfrac{4}{3}\cdot\dfrac{9}{8}\cdot\dfrac{16}{15}\cdot\cdot\cdot\dfrac{2401}{2400}\)
\(A=\)\(\dfrac{2\cdot2}{1\cdot3}\cdot\dfrac{3\cdot3}{2\cdot4}\cdot\dfrac{4\cdot4}{3\cdot5}\cdot\cdot\cdot\dfrac{49\cdot49}{48\cdot50}\)
\(A=\dfrac{\left(2\cdot3\cdot4\cdot49\right)}{\left(1\cdot2\cdot3\cdot48\right)}\cdot\dfrac{\left(2\cdot3\cdot4\cdot\cdot\cdot49\right)}{\left(3\cdot4\cdot5\cdot\cdot\cdot50\right)}\)
\(A=\dfrac{49\cdot2}{50}=\dfrac{98}{50}=\dfrac{49}{35}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
19 tháng 5 2023
a, \(\dfrac{10}{17}\) + \(\dfrac{5}{-13}\) - \(\dfrac{11}{25}\) + \(\dfrac{7}{17}\) - \(\dfrac{8}{13}\)
= ( \(\dfrac{10}{17}\) + \(\dfrac{7}{17}\)) - ( \(\dfrac{5}{13}\) + \(\dfrac{8}{13}\)) - \(\dfrac{11}{25}\)
= \(\dfrac{17}{17}\) - \(\dfrac{13}{13}\) - \(\dfrac{11}{25}\)
= 1 - 1 - \(\dfrac{11}{25}\)
= - \(\dfrac{11}{25}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
19 tháng 5 2023
b, 0,3 - \(\dfrac{93}{7}\) - 70% - \(\dfrac{4}{7}\)
= 0,3 - 0,7 - ( \(\dfrac{93}{7}+\dfrac{4}{7}\))
= - 0,4 - \(\dfrac{97}{7}\)
= - \(\dfrac{2}{5}\) - \(\dfrac{97}{7}\)
= - \(\dfrac{499}{35}\)
HN
19 tháng 7 2016
a) \(\frac{15}{16}\cdot\frac{4}{3}-\frac{1}{2}:\frac{5}{4}+3\)
\(=\frac{5}{4}-\frac{2}{5}+3\)
\(=\frac{25}{20}-\frac{8}{20}+\frac{60}{20}\)
\(=\frac{77}{20}=3\frac{17}{20}\)
b) \(\left(\frac{2}{3}-\frac{3}{8}\right):\left(\frac{3}{5}+\frac{1}{4}\right)\)
\(=\frac{7}{24}:\frac{17}{20}\)
\(=\frac{35}{102}\)
c) \(\frac{15}{8}\cdot\left(\frac{1}{3}+\frac{1}{8}\right)-\frac{3}{8}:\frac{3}{4}\)
\(=\frac{15}{8}\cdot\frac{11}{24}-\frac{1}{2}\)
\(=\frac{55}{64}-\frac{1}{2}\)
\(=\frac{23}{64}\)
d) \(\frac{20}{21}:\left(\frac{4}{5}-\frac{1}{10}\right)+\frac{13}{15}:\frac{5}{26}\)
\(=\frac{20}{21}:\frac{7}{10}+\frac{52}{15}\)
\(=\frac{200}{147}+\frac{52}{15}\)
\(=4\frac{608}{735}\)
DV
0
NN
2
5 tháng 2 2022
A = \(\left(1+\frac{1}{3}\right)\left(1+\frac{1}{8}\right)\left(1+\frac{1}{15}\right)\)\(...\left(1+\frac{1}{2499}\right)\)
A = \(\left(\frac{3}{3}+\frac{1}{3}\right)\left(\frac{8}{8}+\frac{1}{8}\right)\left(\frac{15}{15}+\frac{1}{15}\right)\)\(...\left(\frac{2499}{2499}+\frac{1}{2499}\right)\)
A = \(\frac{4}{3}.\frac{9}{8}.\frac{16}{15}.....\frac{2500}{2499}\)
A = \(\frac{4.9.16.....2500}{3.8.15.....2499}\)
A = \(\frac{\left(2.2\right)\left(3.3\right)\left(4.4\right)...\left(50.50\right)}{3.8.15.24.....2499}\)
A = \(\frac{2.3.4.....50}{3.4.5.6.....51}\)
A = \(\frac{2}{51}\)
Vậy A = \(\frac{2}{51}\)
( Nếu sai mong bạn thông cảm ạ ! )
_HT_
YN
5 tháng 2 2022
Answer:
\(A=\left(1+\frac{1}{3}\right)\left(1+\frac{1}{8}\right)\left(1+\frac{1}{15}\right)...\left(1+\frac{1}{2499}\right)\)
\(=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}...\frac{2500}{2499}\)
\(=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{50^2}{49.51}\)
\(=\frac{2^2.3^2.4^2...50^2}{1.3.2.4.3.5...49.51}\)
\(=\frac{2.50}{51}\)
\(=\frac{100}{51}\)
S
30 tháng 7 2016
Ta có: 1/3 ; 1/15 ; 1/35;...
<=> 1/1.3 ; 1/3.5 ; 1/5.7
=> chữ số thứ 100 là: 1/199.201
Ta có: \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{199.201}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{199}-\frac{1}{201}\)
\(=1-\frac{1}{201}=\frac{200}{201}\)
P
1
XO
9 tháng 8 2020
A = \(\left(1+\frac{1}{3}\right)\left(1+\frac{1}{8}\right)\left(1+\frac{1}{15}\right)...\left(1+\frac{1}{n.n+n.2}\right)=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}...\frac{n^2+2n+1}{n\left(n+2\right)}\)
\(=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{\left(n+1\right)^2}{n\left(n+2\right)}=\frac{2^2.3^2.4^2...\left(n+1\right)^2}{1.3.2.4.3.5...n\left(n+2\right)}=\frac{\left[2.3.4...\left(n+1\right)\right].\left[2.3.4...\left(n+1\right)\right]}{\left(1.2.3...n\right).\left[3.4.5..\left(n+2\right)\right]}\)
\(=\frac{\left(n+1\right).2}{n+2}\)
p/s : giải thích phần n2 + 2n + 1 = (n2 + n) + (n + 1) = n(n + 1) + (n + 1) = (n + 1).(n + 1) = (n + 1)2
Q
1