Đáp án :
Ta có : `AB^2 + AC^2 = BC^2`
`⇔ x^2 + (x + 1)^2 = (x + 2)^2`
`⇔ x^2 + (x + 1) (x + 1) = (x + 2) (x + 2)`
`⇔ x^2 + x (x + 1) + 1 (x + 1) = x (x + 2) + 2 (x + 2)`
`⇔ x^2 + x^2 + x + x + 1 = x^2 + 2x + 2x + 4`
`⇔ x^2 - 2x - 3 = 0`
`⇔ x^2 - 3x + x - 3 = 0`
`⇔ x (x - 3) + (x - 3) = 0`
`⇔ (x - 3) (x + 1) = 0`
`⇔ x - 3 = 0`
Vì `x + 1 > 0`
`⇔ x = 3` `(Đct)`
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