Question

x+(x+1)+(x+2)+...+(x+100)=5353

Answer 1

VT = 101x + 1+2+3+...+100 = 101x + 1/2*100*101

PT <=> 101x + 50*101 = 5353

<=> x + 50 = 53

<=> x = 3

Answer 2

Ta có: x+(x+1)+(x+2)+...+(x+100)=5353

<=>    101x + (1 + 2 + ....+ 100) = 5353

<=>     101x + 2550 = 5353

=> 101x = 5353 - 2550

=> 101x = 2803

=> x =2803 : 101

=> x =????????????

Answer 3

Ta có: x+(x+1)+(x+2)+...+(x+100)=5353

<=>    101x + (1 + 2 + ....+ 100) = 5353

<=>     101x + 5050 = 5353

=> 101x = 5353 - 5050

=> 101x = 303

=> x =303 : 101

=> x = 3

Answer 4

x+(x+1)+(x+2)+...+(x+100)=5353 ( có 100 nhóm )

=> ( x+x+x+....+x) + ( 1+2+3+....+100) = 5353 ( có 101x và 100 số hạng )

=> x101 + (100+1).101:2 = 5353

=> x101 + 5151 = 5353

x101 = 202

=> x = 202 : 101

x = 2

Answer 5

x+(x+1)+...+(x+100) = 5353

101x+ 1+2+3+...+100 = 5353

101x+ (100x101):2=5353

101x+ 5050 = 5353

101x = 5353-5050

101x = 303

x= 303:101=3.

Vậy x=3

Answer 6

<=>\(\left(x+x+x+...+x\right)+\left(1+2+...+100\right)\)\(=5353\)

<=>\(\left(101.x\right)+5050\)\(=5353\)

<=> \(101.x=303\)

=>\(x=\frac{303}{101}\)\(=3\)

Vậy x= 3

Answer 7

Ta có: x+(x+1)+(x+2)+...+(x+100)=5353

<=>    101x + (1 + 2 + ....+ 100) = 5353

<=>     101x + 5050 = 5353

=> 101x = 5353 - 5050

=> 101x = 303

=> x =303 : 101

=> x = 3