`( x + 1 ) + ( x + 2 ) + .... + ( x + 100 ) = 5750`
`⇔ x + 1 + x + 2 + .... + x + 100 = 5750`
`⇔ ( x + x + ... + x ) + ( 1 + 2 + .... + 100 ) = 5750`
`⇔ 100x + ( 100 + 1 ) . 100 : 2 = 5750`
`⇔ 100x + 101 . 50 = 5750`
`⇔ 100x + 5050 = 5750`
`⇔ 100x = 5750 - 5050`
`⇔ 100x = 700`
`⇔ x = 700 : 100`
`⇔ x = 7`
Vậy , `x = 7 .`