1/2 + 1/2^2 + 1/2^3 + 1/2^4 + … + 1/2^9 + 1/2^10 làm sao vậy mấy bạn
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Khách
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T
6
6 tháng 5 2019
có 2 trường hợp
1. 1+2 = 5 + 10 = 15 ( Special )
2. 1+2=3 ( bình thường )
VC
Vũ Cao Minh⁀ᶦᵈᵒᶫ ( ✎﹏IDΣΛ亗 )
CTV
VIP
6 tháng 5 2019
Trả lời
Vì 1 + 2 = 3
Mà 3 = 15
Suy ra 1 + 2 = 15
4 tháng 6 2017
ừ thì giải theo cách lớp 5
\(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\right)\times x=\frac{1}{9}+\frac{2}{8}+...+\frac{9}{1}\)
Gọi vế phải là A
\(A=\frac{10}{9}+\frac{10}{8}+...+\frac{10}{1}-9\)
\(A=10\times\left(\frac{1}{9}+\frac{1}{8}+\frac{1}{7}+...+\frac{1}{2}\right)+1\)
\(A=10\times\left(\frac{1}{10}+\frac{1}{9}+\frac{1}{8}+...+\frac{1}{2}-\frac{1}{10}\right)+1\)
\(A=10\times\left(\frac{1}{10}+\frac{1}{9}+...+\frac{1}{2}\right)-1+1\)
\(\Rightarrow VP=10\times A\)
\(\Rightarrow x=10\)
4 tháng 7 2018
Mình ghi kết quả thôi nhé!
a.2/3
b.1/3
c.3/2
d.18
e.21
f.6/5
g.53/60
h.-13/18
LS
4 tháng 7 2018
giải
a) 1/3 +1/4 +1/12 = 2/3
b) 3/9 + 12/39 - 1/3 = 1/3
c) 72/36 - 1/2 = 3/2
d) 34 - 32 và 1/2 = 18
e) 68/2 - 52/4 = 21
g) 1/2 + 1/3 + 1/4 - 1/5 = 53/60
h) 32/72 - 21/18 = (-13/18)
Mình làm ngắn gọn nhanh nhất rồi đó, rút gon luôn rồi. nha
T
14 tháng 10 2023
\(5\dfrac{9}{10}:\dfrac{3}{2}-\left(2\dfrac{1}{3}\times4\dfrac{1}{2}-2\times2\dfrac{1}{3}\right):\dfrac{7}{4}\)
\(=\dfrac{59}{10}:\dfrac{3}{2}-\left(\dfrac{7}{3}\times\dfrac{9}{2}-2\times\dfrac{7}{3}\right):\dfrac{7}{4}\)
\(=\dfrac{59}{10}\cdot\dfrac{2}{3}-\left[\dfrac{7}{3}\times\left(\dfrac{9}{2}-2\right)\right]:\dfrac{7}{4}\)
\(=\dfrac{59}{15}-\left(\dfrac{7}{3}\times\dfrac{5}{2}\right):\dfrac{7}{4}\)
\(=\dfrac{59}{15}-\dfrac{35}{6}\cdot\dfrac{4}{7}\)
\(=\dfrac{59}{15}-\dfrac{10}{3}\)
\(=\dfrac{59}{15}-\dfrac{50}{15}\)
\(=\dfrac{9}{15}\)
\(=\dfrac{3}{5}\)
\(Toru\)
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) \(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+..........+\frac{1}{200.203}\)
\(=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+........+\frac{3}{200.203}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+......+\frac{1}{200}-\frac{1}{203}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{203}\right)=\frac{1}{3}.\frac{201}{406}=\frac{67}{406}\)
ta nhân 2 lần A lên rồi lấy 2a - A là ra
Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^8}+\frac{1}{2^9}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^8}+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\right)\)
\(A=1-\frac{1}{2^{10}}\)