tình (a b c)^5/a^5 b^5 c^5 thoả mãn a^3 b^3 c^3=3abc

Đặt A=a+b+c5a5+b5+c5.

Ta có: (a + b + c)³ = a³ + b³ + c³ + 6abc + 3(a²b + ab² + b²c + bc² + c²a + a²c).

⇔ (a + b + c)³ = (3abc + 3a²b + 3ab²) + (3abc + 3b²c + 3bc²) + (3abc + 3c²a + 3a²c).

⇔ (a + b + c)³ = 3ab(a + b + c) + 3bc(a + b + c) + 3ac(a + b + c).

⇔ (a + b + c)³ = 3(ab + bc + ca)(a + b + c).

⇔ (a + b + c).[(a + b + c)² – 3(ab + bc + ca)] = 0.

⇔a+b+c=0a+b+c2=3ab+bc+ca    12

Từ (1), ta có: A=a+b+c5a5+b5+c5=05a5+b5+c5=0.

Từ (2), ta có: (a + b + c)² = 3(ab + bc + ca).

⇔ a² + b² + c² + 2ab + 2bc + 2ca = 3ab + 3bc + 3ca.

⇔ a² + b² + c² = ab + bc + ca.

⇔ (a² – 2ab + b²) + (b² – 2bc + c²) + (c² – 2ca + a²) = 0.

⇔ (a – b)² + (b – c)² + (c – a)² = 0.

⇔a=bb=cc=a⇔a=b=c.

Với a = b = c, ta có: A=a+b+c5a5+b5+c5=3a53a5=34.3a53a5=34=81.

Vậy A=0,           khi  a+b+c=081,        khi  a=b=c .