\(\text{Ta có: }\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+.....+\frac{5}{99.100}\)

     \(=5.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\right)\)

       \(=5.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\right)\)

      \(=5.\left(1-\frac{1}{100}\right)\)

   \(=5.\frac{99}{100}\)

      \(=\frac{99}{20}\)

5/1.2 + 5/2.3 + 5/3.4 + ... + 5/99.100

= 5 . ( 1/1.2 + 1/2.3 + 1/3.4 +... + 1/99.100 )

= 5 . ( 1 - 1/2 + 1/2 -1/3 + 1/3 - 1/4 + .... + 1/99 - 1/100 )

= 5 . ( 1 - 1/100 )

= 5 . 99/100

= 99/20

\(\frac{99}{100}\)nhé

tick nha

 S=1.2+ 2.3+.......+99.100 
Nhân cả 2 vế với 3, ta được: 
3S=1.2.3+ 2.3.3+ 3.4.3+ 4.5.3+...... 99.100.3 
= 1.2.3 + 2.3(4-1) + 3.4.(5-2) +...+ 99.100.(101-98) 
= 1.2.3 + 2.3.4 -1.2.3 + 3.4.5-2.3.4 +...+ 99.100.101-98.99.100 
= 99.100.101 
----> S = (99.100.101):3 
S = 333300 
Vậy S=333300 

de

\(=5\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)\)

\(=5.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(=5\left(1-\dfrac{1}{100}\right)\)

\(=5.\dfrac{99}{100}=\dfrac{99}{20}\)

bạn hãy hãy cộng theo cặp cho dễ nhé,có 50 cặp như thế do đó là 101*50=5050

101 ở đâu vậy bn

S=1.2+2.3+3.4+4.5+....+99.100

3S=1.2.3+2.3.3+3.4.3+4.5.3+....+99.100.3

3S=1.2.3+2.3.(4-1)+3.4.(5-1)+4.5(6-3)+....+99.100(101-98)

3S=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-2.4.5+....+99.100.101-98.99.100

3S=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-3.4.5+4.5.6-4.5.6+.......+99.100.101

3S=99.100.101

3S=999900

S=999900:3

S=333300

S=1.2+2.3+3.4+4.5+...+99.100

3S=1.2.3+2.3.(4-1)+3.4.(5-2)+4.5.(6-3)+...+99.100.(101-98)

3S=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+...+99.100.101-98.99.100

3S=99.100.101

S=(99.100.101):3=333300

S=1.2+2.3+3.4+4.5+...+99.100

=>3S=1.2.3+2.3.3+3.4.3+4.5.3+...+99.100.3

=1.2.3+2.3.﴾4‐1﴿+3.4.﴾5‐2﴿+4.5.﴾6‐3﴿+...+99.100.﴾101‐98﴿

=1.2.3+2.3.4‐1.2.3+3.4.5‐2.3.4+4.5.6‐3.4.5+...+99.100.101‐98.99.100

=1.2.3‐1.2.3+2.3.4‐2.3.4+3.4.5‐3.4.5+4.5.6‐4.5.6+...+99.100.101

=99.100.101=999900

=>S=999900:3=333300

Vậy S=333300

C=5/1.2+5/2.3+5/3.4+...+5/99.100

C=5.(1/1.2+1/2.3+1/3.4+...+1/99.100)

C=5.(1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100)

C=5.(1-1/100)

C=5.99/100

C=99/20

K cho mik nha các bạn

\(C=5.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{99.100}\right)\)

     \(=5.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\right)\)

     \(=5.\left(1-\frac{1}{100}\right)\)

     \(=5.\frac{99}{100}=\frac{495}{100}\)

C= 5/1-5/2+5/2-5/3+.....+5/99-5/100

C=5/1-5/100

C=99/20

\(A=5\left(\frac{1}{1.2}+\frac{1}{2.3}+.........+\frac{1}{99.100}\right)\)

\(=5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+............+\frac{1}{99}-\frac{1}{100}\right)\)

\(=5\left(1-\frac{1}{100}\right)\)

\(=5.\frac{99}{100}\)

\(=\frac{99}{20}\)

1. ta có :

\(3^2+4^2=5^{x-1}\)

  \(25=5^{x-1}\)

 \(5^2=5^{x-1}\)

=> x = 3

Ta có : S = 1.2 + 2.3 + 3.4 + ..... + 99.100

=> 3S = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ..... + 99.100.101

=> 3S = 99.100.101

=> S = 99.100.101/3

=> S = 333300 

3A=1.2.3+2.3.(4-1)+3.4.(5-2)+...+99.100.(101-98)

3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100

3A=98.100.101

A=99.100.101 / 3

A=333300

Mình cho bạn dạng tổng quát nha

1.2+2.3+...+n.(n+1)=n(n+1)(n+2) / 3

3A=1.2.3+2.3.(4-1)+...........+99.100.(101-98)

3A=1.2.3+2.3.4-1.2.3+............+99.100.101-98.99.100

3A=99.100.101

A=99.100.101:3

A=333300

ta có 3A=1.2.(3-0)+2.3(4-1)+2.4(5-2)+ ...+97.98(99-96)

            =(1.2.3+2.3.4+3.4.5+....+97.98.99+98.99.100)-(0.1.2+1.2.3+2.3.4+....+96.97.98+97.98.99)

            =98.99.100=970200

=>A=970200:3=323400

mk học bài này rồi