Thực hiện phép tính:
\(A=3.\dfrac{1}{1.2}-5.\dfrac{1}{2.3}+7.\dfrac{1}{3.4}-...+15.\dfrac{1}{7.8}-17.\dfrac{1}{8.9}\)
Chứng minh rằng:
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
Ta có:
\(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}=1-\dfrac{1}{10}=\dfrac{9}{10}\)
Bài 1 : Tính tổng sau
a) \(A=\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}\)
b) \(B=\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+....+\dfrac{1}{23.24}+\dfrac{1}{24.25}\)
c) \(C=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.....+\dfrac{1}{98.99}+\dfrac{1}{99.100}\)
`@` `\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, \(C=\dfrac{99}{100}\)
Tính C:
C=\(\text{}\dfrac{1}{9.10}-\dfrac{1}{8.9}-\dfrac{1}{7.8}-....-\dfrac{1}{2.3}-\dfrac{1}{1.2}\)
Giúp mik với!!!
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
Bài 1: Thực hiện các phép tính:
d) 3,15+2,4=5,55
e) \(\dfrac{5}{7}.\dfrac{2}{11}+\dfrac{5}{7}.\dfrac{9}{11}\)
f) 1,25.3,6+3,6.8,75
h) B= \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)
d, `3,15+2,4=5,55`
e, \(\dfrac{5}{7}.\dfrac{2}{11}+\dfrac{5}{7}.\dfrac{9}{11}=\dfrac{5}{7}\left(\dfrac{2}{11}+\dfrac{9}{11}\right)=\dfrac{5}{7}.\dfrac{11}{11}=\dfrac{5}{7}.1=\dfrac{5}{7}\)
f, `1,25.3,6+3,6.8,75=3,6(1,25+8,75)=3,6.10=36`
\(h,\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}\\ =\dfrac{99}{100}\)
\(e\dfrac{5}{7}\times\left(\dfrac{2}{11}+\dfrac{9}{11}\right)=\dfrac{5}{7}\times1=\dfrac{5}{7}\)
\(f3.6\times\left(1.25+8.75\right)=3.6\times10=36\)
\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=\dfrac{1}{2}-\dfrac{1}{10}\)
\(=\dfrac{2}{5}\)
\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)
\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{10}\)
\(\Rightarrow\dfrac{2}{5}\)
Tính :
a) \(\dfrac{17}{23}.\dfrac{8}{16}.\dfrac{23}{17}.\left(-80\right).\dfrac{3}{4}\)
b) \(\dfrac{5}{11}.\dfrac{18}{29}-\dfrac{5}{11}.\dfrac{8}{29}+\dfrac{5}{11}.\dfrac{19}{29}\)
c) \(\left(\dfrac{13}{23}+\dfrac{1313}{2323}-\dfrac{131313}{232323}\right).\left(\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{7}{12}\right)\)
d) \(\dfrac{1^2}{1.2}.\dfrac{2^{2^{ }}}{2.3}.\dfrac{3^2}{3.4}.\dfrac{4^2}{4.5}.\dfrac{5^2}{5.6}.\dfrac{6^2}{6.7}.\dfrac{7^2}{7.8}.\dfrac{8^2}{8.9}.\dfrac{9^2}{9.10}\)
e) \(\dfrac{2^2}{3}.\dfrac{3^2}{8}.\dfrac{4^2}{15}.\dfrac{5^2}{24}.\dfrac{6^2}{35}\dfrac{7^2}{48}.\dfrac{8^2}{63}.\dfrac{9^2}{80}\)
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!
Tìm x,biết:
(\(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + ........ + \(\dfrac{1}{8.9}\) + \(\dfrac{1}{9.10}\)) . 100 - [ \(\dfrac{5}{2}\) : (\(x+\dfrac{206}{100}\)) ] : \(\dfrac{1}{2}\) = 89
(Dấu . trong bài là dấu nhân ạ)
\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\left(1-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\left(100-10\right)-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow90-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=1\)
\(\Rightarrow\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)=1.2=2\)
\(\Rightarrow\left(x+\dfrac{206}{100}\right)=\dfrac{5}{2}:2=\dfrac{5}{2}.\dfrac{1}{2}=\dfrac{5}{4}\)
\(\Rightarrow x=\dfrac{5}{4}-\dfrac{206}{100}=\dfrac{125}{100}-\dfrac{206}{100}\)
\(\Rightarrow x=-\dfrac{81}{100}\)
1.Tính
A=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.....+\dfrac{1}{99.100}\)
B=\(\dfrac{3}{5.6}+\dfrac{3}{6.7}+\dfrac{3}{7.8}+.....+\dfrac{3}{101.102}\)
C=\(\dfrac{1}{1.2.3}+\dfrac{1}{3.4.5}+\dfrac{1}{5.6.7}\)
D=\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}\)
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: \(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)
Tính hợp lý :
M= \(\dfrac{15}{7.8}\) - \(\dfrac{17}{8.9}\) + \(\dfrac{1}{9.10}\) + \(\dfrac{1}{10.11}\)
\(M=\dfrac{15}{7\cdot8}-\dfrac{17}{8\cdot9}+\dfrac{1}{9\cdot10}+\dfrac{1}{10\cdot11}\)
\(M=\dfrac{1}{8}+\dfrac{1}{7}-\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}\)
\(M=\dfrac{1}{7}-\dfrac{1}{11}=\dfrac{11-7}{7\cdot11}=\dfrac{4}{77}\)
`M=15/(7.8)-17/(8.9)+1/(9.10)+1/(10.11)`
`=1/7-1/8-(1/8-1/9)+1/9-1/10+1/10-1/11`
`=1/7-1/4+2/9-1/11`
`=67/2772`
Theo mình đề bài sai phải là `M=15/(7.8)-17/(8.9)+1/(9.10)+1/(10.11)`
`=1/7-1/8+1/8-1/9+1/9-1/10+1/10-1/11`
`=1/7-1/11=4/77`
M=18+17−18−19+19−110+110−111M=18+17−18−19+19−110+110−111