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`@` `\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, \(C=\dfrac{99}{100}\)
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
1) Tính
a) Ta có: \(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)
Ta có : \(\dfrac{x^2}{5.6}\text{=}\dfrac{x^2}{5}-\dfrac{x^2}{6}\)
\(\dfrac{x^2}{6.7}\text{=}\dfrac{x^2}{6}-\dfrac{x^2}{7}\)
\(...\)
\(\dfrac{x^2}{24.25}\text{=}\dfrac{x^2}{24}-\dfrac{x^2}{25}\)
\(\Rightarrow\) biểu thức chỉ còn :
\(\dfrac{x^2}{5}-\dfrac{x^2}{25}\text{=}\dfrac{5x^2-x^2}{25}\text{=}\dfrac{4x^2}{25}\)
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
đây là tính nhanh à nếu tính bình thường thì tính may tính là ra
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! 
\(M=\dfrac{1}{2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}\)
\(M=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}\)
\(M=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\)
\(M=1-\dfrac{1}{7}\)
\(M=\dfrac{6}{7}\)
tham khảo
https://hoc24.vn/cau-hoi/123134145156167.5003535458609#:~:text=l%C3%BAc%2021%3A02-,1,14,-12.3%2B13.4%2B14.5
vào đi
Ta có:
\(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}=1-\dfrac{1}{10}=\dfrac{9}{10}\)
1/ \(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right).........\left(1-\dfrac{1}{100}\right)\)
\(=\left(\dfrac{2}{2}-\dfrac{1}{2}\right)\left(\dfrac{3}{3}-\dfrac{1}{3}\right).........\left(\dfrac{100}{100}-\dfrac{1}{100}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}...............\dfrac{99}{100}\)
\(=\dfrac{1}{100}\)
2/ \(\dfrac{1}{5.6}+\dfrac{1}{6.7}+.........+\dfrac{1}{99.100}\)
\(=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+........+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\dfrac{1}{5}-\dfrac{1}{100}\)
\(=\dfrac{19}{100}\)
1. \(\left(1-\dfrac{1}{2}\right)\) \(\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)\) \(...\left(1-\dfrac{1}{99}\right)\left(1-\dfrac{1}{100}\right)\)
\(=\left(\dfrac{2}{2}-\dfrac{1}{2}\right)\left(\dfrac{3}{3}-\dfrac{1}{3}\right)\left(\dfrac{4}{4}-\dfrac{1}{4}\right)...\left(\dfrac{99}{99}-\dfrac{1}{99}\right)\left(\dfrac{100}{100}-\dfrac{1}{100}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{98}{99}.\dfrac{99}{100}\)
\(=\dfrac{1}{100}\)
2. \(\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\)
\(=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\dfrac{1}{5}-\dfrac{1}{100}\)
\(=\dfrac{20}{100}\) \(-\dfrac{1}{100}\)
\(=\dfrac{19}{100}\)
a)
\(\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{24.25}\)
\(=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{24}-\dfrac{1}{25}\)
\(=\dfrac{1}{5}-\dfrac{1}{25}\)
\(=\dfrac{4}{25}\)
b)
\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{99.101}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
\(=1-\dfrac{1}{101}\)
\(=\dfrac{100}{101}\)
a) \(\dfrac{1}{5.6}=\dfrac{1}{5}-\dfrac{1}{6}\)
⇒ \(\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{24.25}=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{24}-\dfrac{1}{25}=\dfrac{1}{5}-\dfrac{1}{25}=\dfrac{4}{25}\)b) \(\dfrac{2}{1.3}=1-\dfrac{1}{3}\)
tương tự
\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{99.101}=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}=1-\dfrac{1}{101}=\dfrac{100}{101}\)
B = \(\dfrac{2}{3.4}+\dfrac{2}{4.9}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+\dfrac{2}{7.8}\)
= \(2.\left(\dfrac{1}{3.4}+\dfrac{1}{4.9}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\right)\)
= \(2.\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\right)\)
= \(2.\left(\dfrac{1}{3}-\dfrac{1}{9}-\dfrac{1}{8}\right)\)
= \(2.\dfrac{7}{72}\)
= \(\dfrac{7}{36}\)
A=3.97274066919