Ta có 2x2 + 3x + 2 = \(\dfrac{1}{8}\left(16x^2+24x+16\right)=\dfrac{1}{8}\left(4x+3\right)^2+\dfrac{7}{8}>0\)\(\forall x\)
=> x3 + 2x2 + 3x + 2 > x3
=> y3 > x3 (1)
Lại có y3 = x3 + 2x2 + 3x + 2 = (x + 2)3 - (4x2 + 9x + 6)
Dễ thấy \(4x^2+9x+6=\dfrac{1}{4}\left(4x+\dfrac{9}{2}\right)^2+\dfrac{15}{16}>0\Rightarrow-\left(4x^2+9x+6\right)< 0\)
Khi đó (x + 2)3 - (4x2 + 9x + 6) < (x + 2)3
<=> y3 < (x + 2)3 (2)
Từ (1) ; (2) => x3 < y3 < (x + 2)3
=> y3 = (x + 1)3 (vì y \(\in Z\))
Khi đó x3 + 2x2 + 3x + 2 = (x + 1)3
<=> x3 + 2x2 + 3x + 2 = x3 + 3x2 + 3x + 1
<=> x2 - 1 = 0
<=> (x - 1)(x + 1) = 0
<=> \(x=\pm1\)
x = 1 => y = 2
x = -1 => y = 0
Vậy cặp (x ; y) thỏa mãn : (1 ; 2) ; (-1 ; 0)
