câu 1: tìm x biết: ( \(\dfrac{3}{4.5}\) +\(\dfrac{3}{5.6}\)+\(\dfrac{3}{6.7}\)+\(\dfrac{3}{7.8}\)+\(\dfrac{3}{8.9}\)+\(\dfrac{3}{9.10}\)) . x =\(\dfrac{9}{2}\)
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NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
18 tháng 8 2023
B = \(\dfrac{1}{3.4}\) - \(\dfrac{1}{4.5}\) - \(\dfrac{1}{5.6}\) - \(\dfrac{1}{6.7}\) - \(\dfrac{1}{7.8}\) - \(\dfrac{1}{8.9}\) - \(\dfrac{1}{9.10}\)
B = \(\dfrac{1}{12}\) - ( \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\) + \(\dfrac{1}{7.8}\) + \(\dfrac{1}{8.9}\) + \(\dfrac{1}{9.10}\))
B = \(\dfrac{1}{12}\) - ( \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\) - \(\dfrac{1}{9}\) + \(\dfrac{1}{9}\) - \(\dfrac{1}{10}\))
B = \(\dfrac{1}{12}\) - ( \(\dfrac{1}{4}\) - \(\dfrac{1}{10}\))
B = \(\dfrac{1}{12}\) - \(\dfrac{3}{20}\)
B = - \(\dfrac{1}{15}\)
T
5 tháng 3 2017
đặt \(M=\dfrac{7}{3.4}-\dfrac{9}{4.5}+\dfrac{11}{5.6}-\dfrac{13}{6.7}+\dfrac{15}{7.8}-\dfrac{17}{8.9}+\dfrac{19}{9.10}\)
ta có:
\(M=\dfrac{7}{3.4}-\dfrac{9}{4.5}+\dfrac{11}{5.6}-\dfrac{13}{6.7}+\dfrac{15}{7.8}-\dfrac{17}{8.9}+\dfrac{19}{9.10}\)
\(\Leftrightarrow M=\dfrac{3+4}{3.4}-\dfrac{4+5}{4.5}+\dfrac{5+6}{5.6}-\dfrac{6+7}{6.7}+\dfrac{7+8}{7.8}-\dfrac{8+9}{8.9}+\dfrac{9+10}{9.10}\) \(\Leftrightarrow M=\dfrac{3}{3.4}+\dfrac{4}{3.4}-\dfrac{4}{4.5}-\dfrac{5}{4.5}+\dfrac{5}{5.6}+\dfrac{6}{5.6}-\dfrac{6}{6.7}-\dfrac{7}{6.7}+\dfrac{7}{7.8}+\dfrac{8}{7.8}-\dfrac{8}{8.9}-\dfrac{9}{8.9}+\dfrac{9}{9.10}+\dfrac{10}{9.10}\) \(\Rightarrow M=\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{4}+\dfrac{1}{6}+\dfrac{1}{5}-\dfrac{1}{7}-\dfrac{1}{6}+\dfrac{1}{8}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{8}+\dfrac{1}{10}+\dfrac{1}{9}\) \(\Rightarrow M=\dfrac{1}{3}+\dfrac{1}{10}\)
\(\Rightarrow M=\dfrac{10}{30}+\dfrac{3}{30}\)
\(\Rightarrow M=\dfrac{13}{30}\)
vậy M = \(\dfrac{13}{30}\)
vậy \(\dfrac{7}{3.4}-\dfrac{9}{4.5}+\dfrac{11}{5.6}-\dfrac{13}{6.7}+\dfrac{15}{7.8}-\dfrac{17}{8.9}+\dfrac{19}{9.10}=\dfrac{13}{30}\)
5 tháng 3 2017
\(\dfrac{7}{3.4}-\dfrac{9}{4.5}+\dfrac{11}{5.6}-\dfrac{13}{6.7}+\dfrac{15}{7.8}-\dfrac{17}{8.9}+\dfrac{19}{9.10}=\dfrac{3+4}{3.4}-\dfrac{4+5}{4.5}+\dfrac{5+6}{5.6}-\dfrac{6+7}{6.7}+\dfrac{7+8}{7.8}-\dfrac{8+9}{8.9}+\dfrac{9+10}{9.10}=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}=\dfrac{1}{3}-\dfrac{1}{10}=\dfrac{7}{30}\)
16 tháng 4 2017
Ta có:
\(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}=1-\dfrac{1}{10}=\dfrac{9}{10}\)
16 tháng 7 2023
`@` `\text {Ans}`
`\downarrow`
`a)`
\(A=\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}\)
`=`\(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{8}-\dfrac{1}{9}\)
`=`\(\dfrac{1}{3}-\left(\dfrac{1}{4}-\dfrac{1}{4}\right)-\left(\dfrac{1}{5}-\dfrac{1}{5}\right)-...-\dfrac{1}{9}\)
`=`\(\dfrac{1}{3}-\dfrac{1}{9}\)
`=`\(\dfrac{2}{9}\)
Vậy, \(A=\dfrac{2}{9}\)
`b)`
\(B=\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+...+\dfrac{1}{23\cdot24}+\dfrac{1}{24\cdot25}\)
`=`\(\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{24}-\dfrac{1}{25}\)
`=`\(\dfrac{1}{5}-\left(\dfrac{1}{6}-\dfrac{1}{6}\right)-\left(\dfrac{1}{7}-\dfrac{1}{7}\right)-...-\dfrac{1}{25}\)
`=`\(\dfrac{1}{5}-\dfrac{1}{25}=\dfrac{4}{25}\)
Vậy, \(B=\dfrac{4}{25}\)
`c)`
\(C=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{98\cdot99}+\dfrac{1}{99\cdot100}\)
`=`\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
`=`\(1-\left(\dfrac{1}{2}-\dfrac{1}{2}\right)-\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-...-\dfrac{1}{100}\)
`=`\(1-\dfrac{1}{100}=\dfrac{99}{100}\)
Vậy, \(C=\dfrac{99}{100}\)
5 tháng 3 2018
\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=\dfrac{1}{2}-\dfrac{1}{10}\)
\(=\dfrac{2}{5}\)
14 tháng 3 2017
đây là tính nhanh à nếu tính bình thường thì tính may tính là ra
14 tháng 3 2017
a) 17/23 . 8/16 . 23/17. (-80) . 3/4
= (17/23 . 23/17) . (8/16 . 3/4) . (-80)
= 1 . 3/8 . (-80)
= 3/8 . (-80)
= -30
b) 5/11 . 18/29 - 5/11 . 8/29 + 5/11 . 19/29
= 5/11 . (18/29 - 8/29 + 19/29)
= 5/11 . 1
= 5/11
c)(13/23 + 1313/2323 - 131313/232323).(1/3+1/4 -7/12)
= (13/23 + 1313/2323 - 131313/232323).0
= 0
d) 12/2x2 . 22/2x3 . 32/3x4 . 42/4x5 . 52/5x6 . 62/6x7 . 72/7x8 . 82/8x9 . 92/9x10
= 1/2 . 2/3 . 3/4 . 4/5 . 5/6 . 6/7 . 7/8 . 8/9 .9/10
= 1/10
Khó nhìn quá. Bạn thông cảm nhé! 
23 tháng 3 2021
A=1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100
A=1-1/100 A=99/100 B= (1/5.6+1/6/7+...+1/101.102).3 B=(1/5-1/6+1/6-1/7+...+1/101-1/102).3 B=(1/5-1/102).3 B=97/170
23 tháng 3 2021
1) Tính
a) Ta có: \(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)
3.(1/4.5+1/5.6+...+1/9.10).x=9/2
3.(1/4-1/5+1/5-1/6+...+1/9-1/10).x=9/2
3.(1/4-1/10).x=9/2
3.3/20.x=9/2
9/20.x=9/2
x=9/2:9/20
x=10