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3x/2.5 + 3x/5.8 + 3x/8.11 + 3x/11.14 = 1/21
=> x . ( 3/2.5 + 3/5.8 + 3/8.11 + 3/11.14 ) = 1/21
=> x . ( 1/2.5 + 1/5.8 + 1/8.11 + 1/11.14 ) = 1/21
x . ( 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + 1/11 - 1/14 ) = 1/21
x . ( 1/2 - 1/14 ) = 1/21
x . 3/7 = 1/21
x = 1/21 : 3/7
=> x = 1/9
\(\frac{3x}{2\cdot5}+\frac{3x}{5\cdot8}+\frac{3x}{8\cdot11}+\frac{3x}{11\cdot14}=\frac{1}{21}\)
<=> \(x\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}\right)=\frac{1}{21}\)
<=> \(x\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\right)=\frac{1}{21}\)
<=> \(x\left(\frac{1}{2}-\frac{1}{14}\right)=\frac{1}{21}\)
<=> \(x\cdot\frac{3}{7}=\frac{1}{21}\)
<=> \(x=\frac{1}{9}\)
= 1/3.(1/2-1/5)+1/3.(1/5-1/8)+....+1/3.(1/92-1/95)+1/3.(1/95-1/98)
=1/3.(1/2-1/5+1/5-1/8+....+1/92-1/95+1/95-1/98)
=1/3.(1/2-1/98)
=1/3.24/49
=8/49
Phân tích: 1/2.5 = 1/2 - 1/5
1/5.8 = 1/5 - 1/8
1/8.11 = 1/8 - 1/11
...
1/92.95 = 1/92 - 1/95
1/95.98 = 1/95 - 1/98
Ta có: 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 +...+ 1/92 - 1/95 + 1/95 - 1/98
3 = 3/2.5 + 3/5.8 + 3/8.11 + ...+ 3/92.95 + 3/95.98
3 = 1 - 1/2 + 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 +...+ 1/92 - 1/95 + 1/95 - 1/98
= 1 - 1/98
= 97/98 : 3 = 97/98 x 1/3 = (tự tính)
= 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + .... + 1/29 - 1/32
= 1/2 - 1/32
= ..... ( tự bấm máy tính nhé )
Ta có: \(E=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+....+\frac{1}{29}-\frac{1}{32}\)
\(\Rightarrow E=\frac{1}{2}-\frac{1}{32}\)
\(\Rightarrow E=\frac{16-1}{32}=\frac{15}{32}\)
Vậy \(E=\frac{15}{32}\)
=> 3B = 3.( 1/2.5 + 1/5.8 + 1/8.11 + ........... + 1/122.125)
= 3/2.5 + 3/5.8 + 3/ 8.11 + ......+ 3/122.125
Ta có: 3/ 2.5 = 1/2 - 1/5
3/5.8 = 1/5 -1/8
3/ 8.11 = 1/8 -1/11
..........................
3/122 . 125 = 3/122 - 3/125
=> 3B= 1/2 - 15/5 + 1/5 -1/8 +1/8 - 1/11 +........+1/122 - 1/125
= 1/2 - 1/125 = 125/250 - 2/250= 123/250
=> B= 3B : 3 = 123/250 :3 = 123/250 . 1/3 = 41/250
=> 2C = 2.(1/9.11 + 1/11.13 +....+ 1/97 .99)
= 2/9.11 + 2/11 .13 +.....+ 2/ 97.99
Ta có: 2/9.11 = 1/9 - 1/11
2/11.13 = 2/11 -2/ 13
...............................
2/97.99 = 1/97 - 1/99
=> 2B = 1/9 - 1/11 + 1/11 - 1/13 + ....+ 1/97 - 1/99
= 1/9 -1/99 = 11/99 - 1/99 =10/99
=> B= 2B : B = 10/99 :2 =10/99 . 1/2 = 5/99
Vậy B = 5/99
\(D=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{1979.1982} \)
\(=\frac{1}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{1979.1982}\right)\)
\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{1979}-\frac{1}{1982}\right)\)
\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{1982}\right)\)
\(=\frac{165}{991}\)
S = 1/3 . (1/2 - 1/5 + 1/5 - 1/8 + ... + 1/17 - 1/20)
= 1/3 . (1/2 - 1/20)
= 1/3 . 9/20
= 3/20
\(3S=\frac{5-2}{2.5}+\frac{8-5}{5.8}+\frac{11-8}{8.11}+...+\frac{20-17}{17.20}\)
\(3S=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\)
\(S=\frac{9}{20}:3=\frac{3}{20}\)
\(A=\frac{4}{2.5}+\frac{4}{5.8}+\frac{4}{8.11}+...+\frac{4}{65.68}\)
\(A=\frac{4}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{65.68}\right)\)
\(A=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{65}-\frac{1}{68}\right)\)
\(A=\frac{4}{3}.\left[\frac{1}{2}+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{1}{8}-\frac{1}{8}\right)+...+\left(\frac{1}{65}-\frac{1}{65}\right)-\frac{1}{68}\right]\)
\(A=\frac{4}{3}.\left[\frac{1}{2}-\frac{1}{68}\right]\)
\(A=\frac{4}{3}.\frac{33}{68}\)
\(A=\frac{11}{17}\)
~ Hok tốt ~
\(A=\frac{4}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+....+\frac{1}{65}-\frac{1}{68}\right)\)
\(=\frac{4}{3}\left(\frac{1}{2}-\frac{1}{68}\right)\)
\(=\frac{4}{3}\times\frac{33}{68}=\frac{11}{17}\)
tham khảo ở đây nha
https://olm.vn/hoi-dap/detail/222956295982.html
\(A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{150}-\dfrac{1}{153}\right)\)
\(=\dfrac{1}{3}.\dfrac{151}{306}=\dfrac{151}{918}\)
nay tui con thay
\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{98.101}\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{98}-\frac{1}{101}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{101}\right)\)
\(=\frac{1}{3}\cdot\frac{99}{202}=\frac{33}{202}\)