kinh q u e

1 số ko thể tưởng tượng

\(A=20\times21+21\times22+...+99\times100\)

\(3\times A=20\times21\times\left(22-19\right)+21\times22\times\left(23-20\right)+...+99\times100\times\left(101-98\right)\)

\(=20\times21\times22-19\times20\times21+...+99\times100\times101-98\times99\times100\)

\(=99\times100\times101-19\times20\times21\)

Suy ra \(A=\frac{99\times100\times101-19\times20\times21}{3}=360640\)

\(B=3\times4\times5+4\times5\times6+...+98\times99\times100\)

\(4\times B=3\times4\times5\times\left(6-2\right)+4\times5\times6\times\left(7-3\right)+...+98\times99\times100\times\left(101-97\right)\)

\(=3\times4\times5\times6-2\times3\times4\times5+...+98\times99\times100\times101-97\times98\times99\times100\)

\(=98\times99\times100\times101-2\times3\times4\times5\)

Suy ra \(B=\frac{98\times99\times100\times101-2\times3\times4\times5}{4}=24497520\)

47. (23 + 50)- 23. ( 47+50 )

= 47. 23+ 47. 50- 23. 47 - 23. 50

=47. 50 - 23. 50

= 50 .( 47 -23)

=50. 24 = 1200

(-33 ). (-5) - 33 . 4 + 33

= 33 ( 5 - 4 + 1)

=33 . 2 = 66

giúp mình trả lời câu hỏi này với 

=1x-2x3x-4x...x99/2x3x4x5x6x...x100=-1/100

Số lượng số hạng x:

(99-1):2 + 1= 50 (số hạng)

(x+1) + (x+3) + (x+5) +...+ (x+99)= 2450

50x + (1+99) x 25 = 2450 (chia thành 25 cặp)

50x + 2500 = 2450

50x= 2450 - 2500

50x = -50

x= -50:50

x=-1

(x+1)+(x+3)+...+(x+99)=2450

=>99x+\(\dfrac{\left(99-1\right)+1\cdot\left(1+99\right)}{2}\)=2450

=>99x+4950=2450

=>99x=2450-4950=-2500

=>x=-2500:99=\(\dfrac{-2500}{99}\).

cau 1 :

Tu dau bai ta co:

 X x 50 + ( 100 + 2 ) x 50 : 2 = 2700

 X x 50 + 2550 = 2700

X x 50 = 2700 - 2550

X x 50 = 150

X = 150 : 50

X = 3

Bai 2 

1 ) ( 99 + 0 ) x 100 : 2 = 4950

2 ) ( 99 + 1 ) x 99 : 2 = 4950

3 )    day tren co so cac so hang la :

              ( 99 - 1 ) : 2 + 1 = 50 ( so )

Vay tong la : ( 99 + 1 ) x 50 : 2 = 2500

lik e cho minh nha moi nguoi

\(\left(x+1\right)+\left(x+3\right)+\left(x+5\right)+...+\left(x+99\right)=0\\ \left(x+x+...+x\right)+\left(1+3+5+...+99\right)=0\\ 50x+2500=0\\ x=-50\)

Vậy x = -50

Ta có: \(\left(x+1\right)+\left(x+3\right)+...+\left(x+99\right)=0\)

\(\Leftrightarrow50x=-2500\)

hay x=-50

`(x+1)+(x+3)+(x+5)+.....+(x+99)=0`

`=>(x+x+.....+x)+(1+3+5+......+99)=0`

`=>50x+(100.50)/2=0` (do từ 1 đến 99 có 50 số nên vì vậy cũng có 50x)

`=>50x+50.50=0`

`=>50(x+50)=0`

`=>x+50=0`

`=>x=-50`

Vậy `x=-50`

(x+1)+(x+3)+(x+5)+...+(x+99)=0

<=> 50x + (1+99) x 25=0

<=>50x+ 2500=0

<=>50x= -2500

<=>x= -50 

=>S={-50}