Đáp án + Giải thích các bước giải:

Đặt B = 1.99+2.98+...+99.1

=1.99+2.(99-1)+3.(99-2)+...+98.(99-97)+99.(99-98)

=1.99+2.99-1.2+3.99-2.3+...+98.99-97.98+99.99-98.99 

=(1.99+2.99+3.99+...+98.99+99.99)-(1.2+2.3+3.4+...+98.99) - 99.100 + 99.100 ( thêm bớt 99.100)

=99.(1+2+...+99+100)-(1.2+2.3+...+98.99+99.100) 

=99.5050-(1.2+2.3+...+98.99+99.100)

=499950-(1.2+2.3+...+98.99+99.100)

khi đó A = $\frac{499950-(1.2+2.3+...+98.99+99.100)}{1.2+2.3+...+98.99+99.100}$ 

=$\frac{499950}{1.2+2.3+...+98.99+99.100}$ - $\frac{1.2+2.3+...+98.99+99.100}{1.2+2.3+...+98.99+99.100}$ 

=$\frac{499950}{1.2+2.3+...+98.99+99.100}$ - 1 (*)

đặt C = 1.2+2.3+...+98.99+99.100

⇒3C = 1.2.3+2.3.3+....+99.100.3

⇒3C =1.2. ( 3 - 0 ) +2.3.( 4 - 1 ) + .....+ 99.100 .( 101 -98 )

⇒3C = 1.2. 3 - 0.1.2+ 2 . 3 . 4 - 1 . 2 . 3+ ... +  99 . 100.101 -  98 . 99.100

⇒3C=99 .100 .101 

⇒3C =999900

⇒C = 333300

thay vào (*) ta có 

A =$\frac{499950}{333300}$ 

⇔A = $\frac{3}{2}$ 

Nếu mà muốn xóa bài mik thì ns với mik sai ở đâu nha .

Đáp án:

 `(1.99+2*98+3*97+...+99*1)/(1.2+2.3+3.4+...+99.100)`

Xét 

`1.99+2*(99-1) +3(99-2)+...+99*(99-98)`

`=1*99+2*99+3*99+...+99*99-1*2-2*3-3*4-...-98*99`

`=99(1+2+3+...+99)-(1*2+2*3+3*4-...+98*99)`

`99*4950-33*98*100`

xét` C=1*2+2*3+3*4-...+98*99`

`3C = 1 . 2 . 3 + 2 . 3 . 3 + 3 . 4 . 3 + 4 . 5 . 3 + ... + 98 . 99 . 3`

`3C = 1 . 2 . 3 + 2 . 3 . ( 4 - 1 ) + 3 . 4 . ( 5 - 2 ) + 4 . 5 . ( 6 - 3 ) + ... + 98 . 99 . ( 100 - 97 )`

`3C = 1 . 2 . 3 + 2 . 3 . 4 - 1 . 2 . 3 + 3 . 4 . 5 - 2 . 3 . 4 + 4 . 5 . 6 - 3 . 4 . 5 + ... + 98 . 99 . 100 - 97 . 98 . 99`

`3C= 98 . 99 . 100`

`=>C=33*98*100`

`B=1.2+2.3+3.4+...+99.100`

`3B= 1.2.3+2.3.4+3.4.3 +......+ 99.100.3`

`3B= 1.2. (3 - 0) + 2.3.(4 - 1) +3.4. (5 - 2)....... . 99.100. (101 - 98)`

`3B = (1.2.3 + 2.3.4 + 3.4.5 +...... + 99.100.101) - (0.1.2 + 1.2.3 + 2.3.4 +.......+ 98.99.100)`

`3B = 99.100.101 - 0.1.2`

`3B= 99*100*101`

`B=33*100*101`

`=>A=(99*4950 +33*98*100)/(33*100*101)`

`A=14850/(100*101)+98/101`

Giải thích các bước giải: