=1/10 - 1/11+......+1/99-1/100

=1/10-1/100

=9/100

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

a)N=1.2+2.3+3.4+4.5+......+99.100

3N=1.2.3+2.3.3+3.3.4+3.4.5+......+3.99.100

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=999900

N=333300

b)P=10.11+11.12+12.13+....+29.30

3P=10.11.3+11.12.3+....+29.30.3

3P=10.11.12-9.10.11+11.12.13-10.11.12+.........+29.30.31-28.29.30

3P=29.30.31-9.10.11=25980

P=8660

R=1²+2²+3²+....+100²

R=1.(2-1)+2.(3-1)+.........+100.(101-1)

R=1.2-1+2.3-2+3.4-3+.....+100.101-100

R=(1.2+2.3+3.4+..........+100.101)-(1+2+3+.........+99+100)

\(R=\left(N+100.101\right)-\frac{100.101}{2}\)

R=343400-5050=338350

S=100.1+2.(100-1)+3.(100-2)+......+99.(100-98)+100(100-99)

S=100.1+100.2-1.2+100.3-2.3+.......+99.100-98.99+100.100-99.100

S=100(1+2+...+100)-(1.2+2.3+....+99.100)

S=100.(100.101:2)-N

S=505000-333300=171700

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

chịuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu

\(\dfrac{x}{10.11}\) + \(\dfrac{x}{11.12}\) +................+ \(\dfrac{x}{99.100}\)= \(\dfrac{99}{100}\)

\(x\)( \(\dfrac{1}{10.11}+\dfrac{1}{11.12}+\dfrac{1}{12.13}\) +..........+\(\dfrac{1}{99.100}\)) = \(\dfrac{99}{100}\)

\(x\). ( \(\dfrac{1}{10}\) - \(\dfrac{1}{11}\) + \(\dfrac{1}{11}\) - \(\dfrac{1}{12}\) + \(\dfrac{1}{12}\) - \(\dfrac{1}{13}\)+...........+\(\dfrac{1}{99}\)- \(\dfrac{1}{100}\)) = \(\dfrac{99}{100}\)

\(x\). \(\dfrac{9}{100}\)           = \(\dfrac{99}{100}\)

\(x\)     = \(\dfrac{99}{100}\) : \(\dfrac{9}{100}\)

\(x\)    = 11

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:

$3S=10.11(12-9)+11.12(13-10)+12.13(14-11)+...+98.99(100-97)+99.100(101-98)$

$=(10.11.12+11.12.13+12.13.14+...+98.99.100+99.100.101)-(9.10.11+10.11.12+...+97.98.99+98.99.100)$

$=99.100.101-9.10.11$

$\Rightarrow S=\frac{99.100.101-9.10.11}{3}=33.100.101-3.10.11$