1-\(\dfrac{1}{1.2}\)-\(\dfrac{1}{2.3}\)-...-\(\dfrac{1}{9.10}\)<\(\dfrac{1}{9}\)
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2
VQ
1 tháng 4 2023
\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+...+\(\dfrac{1}{9.10}\)< 1
Ta có:
\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+...+\(\dfrac{1}{9.10}\)
= 1 - \(\dfrac{1}{2}\)+\(\dfrac{1}{2}\) - \(\dfrac{1}{3}\)+...+\(\dfrac{1}{9}\) - \(\dfrac{1}{10}\)
=1 - \(\dfrac{1}{10}\)
=\(\dfrac{9}{10}\)
=>\(\dfrac{9}{10}\) < 1
Vậy \(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+...+\(\dfrac{1}{9.10}\) < 1
KV
15 tháng 9 2023
C = 1/(9.10) - 1/(8.9) - 1/(7.8) - ... - 1/(2.3) - 1/(1.2)
= 1/9 - 1/10 - 1/8 + 1/9 - 1/7 + 1/8 - ... - 1/2 + 1/3 - 1 + 1/2
= 1/9 - 1/10 + 1/9 - 1
= 2/9 - 11/10
= -79/90
16 tháng 4 2017
Ta có:
\(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}=1-\dfrac{1}{10}=\dfrac{9}{10}\)
21 tháng 7 2023
\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\left(1-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\left(100-10\right)-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow90-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=1\)
\(\Rightarrow\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)=1.2=2\)
\(\Rightarrow\left(x+\dfrac{206}{100}\right)=\dfrac{5}{2}:2=\dfrac{5}{2}.\dfrac{1}{2}=\dfrac{5}{4}\)
\(\Rightarrow x=\dfrac{5}{4}-\dfrac{206}{100}=\dfrac{125}{100}-\dfrac{206}{100}\)
\(\Rightarrow x=-\dfrac{81}{100}\)
EJ
14 tháng 8 2017
a, \(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]=178\)
\(\left(1-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]=178\)
\(\dfrac{9}{10}.100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]=178\)
\(90-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]=178\)
\(\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\) \(=-88\)
\(x+\dfrac{206}{100}=\dfrac{-5}{176}\)
\(x=\dfrac{-5}{176}-\dfrac{206}{100}\)
\(x=\dfrac{-9198}{4400}\)
JP
14 tháng 8 2017
a) \(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\left(1-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\dfrac{9}{10}.100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(90-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=90-89\)
\(\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=1\)
\(\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)=\dfrac{1}{2}\)
\(x+\dfrac{206}{100}=5\)
\(x=5-\dfrac{206}{100}\)
\(x=\dfrac{147}{50}\)
Vậy \(x=\dfrac{147}{50}\)
10 tháng 6 2017
1)Tính
a)\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+..........+\dfrac{1}{9.10}\)
=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.....+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=1-\dfrac{1}{10}\)
\(=\dfrac{9}{10}\)
b)\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.........+\dfrac{1}{99.100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+..............+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}\)
\(=\dfrac{99}{100}\)
2) tìm x
\(a\)) \(\dfrac{2}{5}+\dfrac{4}{5}x-\dfrac{7}{5}\)\(=\dfrac{9}{5}\)
\(\dfrac{4}{5}x+\dfrac{7}{5}=\dfrac{9}{5}-\dfrac{2}{5}\)
\(\dfrac{4}{5}x+\dfrac{7}{5}=\dfrac{7}{5}\)
\(\dfrac{4}{5}x=\dfrac{7}{5}-\dfrac{7}{5}\)
\(\dfrac{4}{5}x=0\)
\(x=0:\dfrac{4}{5}\)
\(x=0\)
b)\(\dfrac{2}{5}x-\dfrac{6}{4}=\dfrac{8}{5}\)
\(\dfrac{2}{5}x=\dfrac{8}{5}+\dfrac{6}{4}\)
\(\dfrac{2}{5}x=\dfrac{31}{10}\)
\(x=\dfrac{31}{10}:\dfrac{2}{5}\)
\(x=\dfrac{31}{4}\)
10 tháng 6 2017
1. Tính:
a. \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{9.10}\)
= \(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
= \(\dfrac{1}{1}-\dfrac{1}{10}\)
= \(\dfrac{10}{10}-\dfrac{1}{10}\)
= \(\dfrac{9}{10}\)
b. \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
= \(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
= \(\dfrac{1}{1}-\dfrac{1}{100}\)
= \(\dfrac{100}{100}-\dfrac{1}{100}\)
= \(\dfrac{99}{100}\)
2. Tìm x, biết:
a. \(\dfrac{2}{5}+\dfrac{4}{5}x-\dfrac{7}{5}=\dfrac{9}{5}\)
\(\dfrac{4}{5}x-\dfrac{7}{5}=\dfrac{9}{5}-\dfrac{2}{5}\)
\(\dfrac{4}{5}x-\dfrac{7}{5}=\dfrac{7}{5}\)
\(\dfrac{4}{5}x=\dfrac{7}{5}+\dfrac{7}{5}\)
\(\dfrac{4}{5}x=\dfrac{14}{5}\)
\(x=\dfrac{14}{5}:\dfrac{4}{5}\)
\(x=\dfrac{14}{5}.\dfrac{5}{4}\)
\(x=14.\dfrac{1}{4}\)
\(x=\dfrac{14}{4}\)
Vậy \(x=\dfrac{14}{4}\)
b. \(\dfrac{2}{5}x-\dfrac{6}{4}=\dfrac{8}{5}\)
\(\dfrac{2}{5}x=\dfrac{8}{5}+\dfrac{6}{4}\)
\(\dfrac{2}{5}x=\dfrac{32}{20}+\dfrac{30}{20}\)
\(\dfrac{2}{5}x=\dfrac{62}{20}\)
\(\dfrac{2}{5}x=\dfrac{31}{10}\)
\(x=\dfrac{31}{10}:\dfrac{2}{5}\)
\(x=\dfrac{31}{10}.\dfrac{5}{2}\)
\(x=\dfrac{31}{2}.\dfrac{2}{2}\)
\(x=\dfrac{31}{2}.1\)
\(x=\dfrac{31}{2}\)
Vậy \(x=\dfrac{31}{2}\)
bài này mk tự làm ko sao chép trên mạng
nếu thấy đúng thì tick đúng cho mk nha
NH
10 tháng 3 2023
a)
`1/1-1/2`
`=2/2-1/2`
`=1/2`
b)
`1/(1*2)+1/(2*3)`
`=1/1-1/2+1/2-1/3`
`=1/1-1/3`
`=3/3-1/3`
`=2/3`
c)
\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =\dfrac{1}{1}-\dfrac{1}{100}\\ =\dfrac{99}{100}\)
d)
\(\dfrac{3}{1\cdot2}+\dfrac{3}{2\cdot3}+...+\dfrac{3}{99\cdot100}\) đề phải như thế này chứ nhỉ?
\(=\dfrac{1\cdot3}{1\cdot2}+\dfrac{1\cdot3}{2\cdot3}+...+\dfrac{1\cdot3}{99\cdot100}\\ =3\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\right)\\ =3\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =3\left(\dfrac{1}{1}-\dfrac{1}{100}\right)\\ =3\cdot\dfrac{99}{100}\\ =\dfrac{297}{100}\)
\(1-\dfrac{1}{1.2}-\dfrac{1}{2.3}-...-\dfrac{1}{9.10}\\ =1-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{9.10}\right)\\ =1-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\\ =1-\left(1-\dfrac{1}{10}\right)\\ =1-\dfrac{9}{10}\\ =\dfrac{1}{10}\\ Vì\dfrac{1}{10}< \dfrac{1}{9}\\ Vậy1-\dfrac{1}{1.2}-\dfrac{1}{2.3}-...-\dfrac{1}{9.10}< \dfrac{1}{9}\)
=1-(1-1/2+1/2-1/3+...+1/9-1/10)
=1-1+1/10
=1/10<1/9