D = 1/2.5 + 1/5.8 + 1/8.11 +.... + 1/1979.1982.
NhỜ MọI NgƯờI GiÚp NhÉ.
D=1/2.5+1/5.8+1/8.11+...+1/1979.1982
`D=1/2.5+1/5.8+1/8.11+...+1/1979.1982`
`=3/3(1/2.5+1/5.8+1/8.11+...+1/1979.1982)`
`=1/3(3/2.5+3/5.8+3/8.11+...+3/1979.1982)`
`=1/3(1/2-1/5+1/5-1/8+1/8-1/11+...+1/1979-1/1982)`
`=1/3(1/2-1/1982)`
`=1/3(991/1982-1/1982)`
`=1/3 . 495/991`
`=165/991`
1/2.5 + 1/5.8 + 1/ 8.11 + ... +1/98.11 = ?
Giúp mình nha , gấp lắm!
\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{98.101}\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{98}-\frac{1}{101}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{101}\right)\)
\(=\frac{1}{3}\cdot\frac{99}{202}=\frac{33}{202}\)
sửa lại đề 1 chút :
\(\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{98\cdot101}\)
\(=\frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{98\cdot101}\right)\)
\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{98}-\frac{1}{101}\right)\)
\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{101}\right)\)
\(=\frac{1}{3}\cdot\frac{99}{202}\)
\(=\frac{33}{202}\)
A= 1/2.5 + 1/5.8 + 1/8.11 + .....+ 1/150.153
tham khảo ở đây nha
https://olm.vn/hoi-dap/detail/222956295982.html
\(A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{150}-\dfrac{1}{153}\right)\)
\(=\dfrac{1}{3}.\dfrac{151}{306}=\dfrac{151}{918}\)
b=1/2.5+1/5.8+1/8.11+...+1/92.95
A = 1/2.5 + 1/5.8 + 1/8.11 + ... + 1/92.95 + 1/95.98
A = 1/3 . ( 3/2.5 + 3/5.8 + 3/8.11 + ... + 3/92.95 + 3/95.98 )
A = 1/3 . ( 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + ... + 1/92 - 1/95 + 1/95 - 1/98 )
A = 1/3 . ( 1/2 - 1/98 )
A = 1/3 . 24/49
A = 8/49 tick cho tui
Chứng minh rằng với mọi số tư nhiên n khác 0 ta đều có:1/2.5+1/5.8+1/8.11+...+1/(3n-1)(3n+2)=n/6n+4
Đặt A=1/2.5+1/5.8+...+1/(3n-1)(3n+2)
3A=3/2.5+3/5.8+....+3/(3n-1)(3n+2)
3A=1/2-1/5+1/5-1/8+....+1/3n-1-1/3n+2
3A=1/2-1/3n+2
3A=3n/6n+4
A=(3n/6n+4) /3
A=n/6n+4(đpcm)
Tính giá trị biểu thức
A=\(\dfrac{1}{2.5}\)+\(\dfrac{1}{5.8}\)+\(\dfrac{1}{8.11}\)+.....+\(\dfrac{1}{92.95}\)+\(\dfrac{1}{95.98}\)
Mong mn giúp đỡ
1/2.5 - 1/5.8 -1/8.11 - ...- 1/302.305
=1/3.3(1/2.5-1/5.8-1/8.11-...-1/302.305)
=1/3.(3/2.5-3/5.8-3/8.11-...-3/302.305)
=1/3(1/2-1/5-1/5-1/8-1/8-1/11-...-1/302-1/305)
=1/3[(1/2-1/305)+(1/5-1/5)+...+(1/302-1/302)
=1/3*(1/2-1/305)=1/3*(305/610-1/610)=1/3*304/610=152/915
hình như mình làm sai hoặc sai đề , sao số lớn ghê
1/2.5+1/5.8+1/8.11+....+1/29.32
1/2.5+1/5.8+1/8.11+...+1/29.32 (khoảng cách từ 2-5;5-8;8-11;...;29-32 là 3) suy ra
=1/3(1/2-1/5+1/5-1/8+1/8-1/11+...+1/29-1/32) (-1/5+1/5;-1/8+1/8;-1/11+1/11=0) suy ra =1/3(1/2-1/32)=1/3.15/32=5/32
1/2.5+1/5.8+1/8.9+............+1/29.32
=1/2-1/5+1/5-1/8+...............+1/29-1/32
=1/2-1/32
=15/32
ai tích mk=>mk tích lại
1/2.5+1/5.8+1/8.11+...+ 1/47.50 = ?