x/10.11 + x/11.12 +...+ x/99.100=99/100.
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ta có : 1.2+2.3+3.4+.....+99.100=99.100.101 /3 =333300
mà 1.2+2.3+....+9.10+9.10.11/3=330
=>E= 333300-330=332970
Đặt : \(A=10.11+11.12+...+98.99+99.100\)
\(\Rightarrow3A=10.11.3+11.12.3+...+98.99.3+99.100.3\)
\(\Rightarrow3A=10.11.\left(12-9\right)+11.12.\left(13-10\right)+...+\)\(98.99.\left(100-97\right)+99.100.\left(101-98\right)\)
Lời giải:
$3S=10.11(12-9)+11.12(13-10)+12.13(14-11)+...+98.99(100-97)+99.100(101-98)$
$=(10.11.12+11.12.13+12.13.14+...+98.99.100+99.100.101)-(9.10.11+10.11.12+...+97.98.99+98.99.100)$
$=99.100.101-9.10.11$
$\Rightarrow S=\frac{99.100.101-9.10.11}{3}=33.100.101-3.10.11$
Số số hạng là(99,100-10,11):0,01+1=8900
=>Tổng=8900:2x(10,11+99,100)=485984,5
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!
\(S=9\cdot10+10\cdot11+11\cdot12+...+99\cdot100\)
\(3S=9\cdot10\cdot3+10\cdot11\cdot3+11\cdot12\cdot3+...+99\cdot100\cdot3\)
\(3S=9\cdot10\cdot\left(11-8\right)+10\cdot11\cdot\left(12-9\right)+...+99\cdot100\cdot\left(101-98\right)\)
\(3S=9\cdot10\cdot11-8\cdot9\cdot10+10\cdot11\cdot12-9\cdot10\cdot11+...+99\cdot100\cdot101-98\cdot99\cdot100\)
\(3S=99\cdot100\cdot101\)
\(S=\frac{99\cdot100\cdot101}{3}=333300\)
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
\(\dfrac{x}{10.11}\) + \(\dfrac{x}{11.12}\) +................+ \(\dfrac{x}{99.100}\)= \(\dfrac{99}{100}\)
\(x\)( \(\dfrac{1}{10.11}+\dfrac{1}{11.12}+\dfrac{1}{12.13}\) +..........+\(\dfrac{1}{99.100}\)) = \(\dfrac{99}{100}\)
\(x\). ( \(\dfrac{1}{10}\) - \(\dfrac{1}{11}\) + \(\dfrac{1}{11}\) - \(\dfrac{1}{12}\) + \(\dfrac{1}{12}\) - \(\dfrac{1}{13}\)+...........+\(\dfrac{1}{99}\)- \(\dfrac{1}{100}\)) = \(\dfrac{99}{100}\)
\(x\). \(\dfrac{9}{100}\) = \(\dfrac{99}{100}\)
\(x\) = \(\dfrac{99}{100}\) : \(\dfrac{9}{100}\)
\(x\) = 11