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Ta có A=1/10.11+1/11.12+...+1/98.99+1/99.100
=1/10-1/11+1/11-1/12+...+1/98-1/99+1/99-1/100
=1/10-1/100
=10/100-1/100
=9/100
Vậy A=9/100
Giải:
A=1/10.11+1/11.12+...+1/98.99+1/99.100
A=1/10-1/11+1/11-1/12+...+1/98-1/99+1/99-1/100
A=1/10-1/100
A=9/100
Chúc bạn học tốt!
10.11+11.12+12.13+...+97.98+98.99+99.100
=10-11+11-12+12-13+...+97-98+98-99+99-100
=10-100
=-90
Đặt A = 10.11 + 11.12 + ... + 98.99 + 99.100
3A = 10.11.3 + 11.12.3 + ... + 98.99.3 + 99.100.3
3A = 10.11.(12 -9) + 11.12.(13-10) + ... + 98.99.(100 - 97) + 99.100.(101-98)
3A = 10.11.12 - 9.10.11 + 11.12.13 - 10.11.12 + ... + 98.99.100 - 97.98.99 + 99.100.101 - 98.99.100
3A = (10.11.12 + 11.12.13 + ... + 98.99.100 + 99.100.101) - (9.10.11 + 10.11.12 + ... + 97.98.99 + 98.99.100)
3A = 99.100.101 - 9.10.11
3A = 999799
A = 999799 : 3
3N = 1.2.3+2.3(4-1)+3.4.(5-2)+.+99.100.(101-98)
3N = 1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+.+99.100.101-98.99.100
3N = 99.100.101
3N=33.100.101=333300
b)
tổng này có 99-10+1=90 (số hạng):
10,11 + 11,12 + 12,13 +............+ 98,99 + 99,100 =
10,100 + 11,11 + 12,12 + .......... + 98,98 + 99,99 =
(10,10 + 99,99) x 90 : 2 = 4954,05
c)
R=1.(2-1)+2.(3-1)+.....+100.(101-1)
=1.2-1.1+2.3-1.2+......+100.101-1.100
=(1.2+2.3+.....+99.100+100.101)-(1+2+3+...+100)
=[1.2.3+2.3.(4-1)+........100.101.(102-99)]:3+[(100+1).100:2]
(tổng trên chia cho 3 nên cuối cùng chia 3)
=(1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+.....100.101.102-99.100.101):3+5050
=(100.101.102) :3 +5050
=348450
d)=1.100+2.(100-1)+.....+100.(100-99)
=1.100+2.100-1.2+3.100-2.3+........+100.100-99.100
=100.(1+2+3+.......+100)-(1.2+2.3+3.4+....+99.100)
=100.\(\frac{101.100}{2}-\frac{99.100.101}{3}\) =505000-333300=171700
p/s mỏi tay, bấm mình nhé
\(\dfrac{x}{10.11}\) + \(\dfrac{x}{11.12}\) +................+ \(\dfrac{x}{99.100}\)= \(\dfrac{99}{100}\)
\(x\)( \(\dfrac{1}{10.11}+\dfrac{1}{11.12}+\dfrac{1}{12.13}\) +..........+\(\dfrac{1}{99.100}\)) = \(\dfrac{99}{100}\)
\(x\). ( \(\dfrac{1}{10}\) - \(\dfrac{1}{11}\) + \(\dfrac{1}{11}\) - \(\dfrac{1}{12}\) + \(\dfrac{1}{12}\) - \(\dfrac{1}{13}\)+...........+\(\dfrac{1}{99}\)- \(\dfrac{1}{100}\)) = \(\dfrac{99}{100}\)
\(x\). \(\dfrac{9}{100}\) = \(\dfrac{99}{100}\)
\(x\) = \(\dfrac{99}{100}\) : \(\dfrac{9}{100}\)
\(x\) = 11
Lời giải:
$3S=10.11(12-9)+11.12(13-10)+12.13(14-11)+...+98.99(100-97)+99.100(101-98)$
$=(10.11.12+11.12.13+12.13.14+...+98.99.100+99.100.101)-(9.10.11+10.11.12+...+97.98.99+98.99.100)$
$=99.100.101-9.10.11$
$\Rightarrow S=\frac{99.100.101-9.10.11}{3}=33.100.101-3.10.11$
\(S=9\cdot10+10\cdot11+11\cdot12+...+99\cdot100\)
\(3S=9\cdot10\cdot3+10\cdot11\cdot3+11\cdot12\cdot3+...+99\cdot100\cdot3\)
\(3S=9\cdot10\cdot\left(11-8\right)+10\cdot11\cdot\left(12-9\right)+...+99\cdot100\cdot\left(101-98\right)\)
\(3S=9\cdot10\cdot11-8\cdot9\cdot10+10\cdot11\cdot12-9\cdot10\cdot11+...+99\cdot100\cdot101-98\cdot99\cdot100\)
\(3S=99\cdot100\cdot101\)
\(S=\frac{99\cdot100\cdot101}{3}=333300\)
co \(\frac{1}{9\cdot10}=\frac{1}{9}-\frac{1}{10}\)
\(\frac{1}{10\cdot11}=\frac{1}{10}-\frac{1}{11}\)
............
\(\frac{1}{x\left(x+1\right)}=\frac{1}{x}-\frac{1}{x+1}\)
nen \(\frac{1}{9\cdot10}+\frac{1}{10\cdot11}+...+\frac{1}{x\left(x+1\right)}\)
\(=\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}-...+\frac{1}{x}-\frac{1}{x+1}\)
=\(\frac{1}{9}-\frac{1}{x+1}\)
2 . ( \(\frac{1}{9\cdot10}+\frac{1}{10\cdot11}+...+\frac{1}{x\left(x+1\right)}\))
= 2 . ( \(\frac{1}{9}-\frac{1}{x+1}\)) = \(\frac{2}{9}-\frac{2}{x+1}\)
MÌNH BIK LÀM CÂU A THUI, mình ko ghi lại đề nha
P=1/2.2/3.3/4........99/100
(Nhân tử với tử, mẫu nhân với mẫu ) ta có
P=1.2.3.4.......99/2.3.4...........100
P=1/100
đặt a=1/10.11+ 1/11.12+..+1/99.100
=1/10-1/11+1/11-1/12+...+1/99-1/100
=1/10-1/100=9/100
vậy a=9/100
9/100 lấy 1/10-1/100 là ra