Bài 11 : Ta có :
a^3 + b^3 + c^3 = 3abc
<=> a^3 + b^3 + c^3 - 3abc = 0
<=> (a+b)^3 + c^3 - 3ab(a+b+c) = 0
<=> (a+b+c)[(a+b)^2 - (a+b).c + c^2 ] - 3ab(a+b+c) = 0
<=> (a+b+c)[(a+b)^2 - ac - bc + c^2 - 3ab] = 0
<=> (a+b+c)[a^2 + 2ab + b^2 - ac - bc + c^2 - 3ab] = 0
<=> (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ac) = 0
TH 1 : a + b + c = 0
=> a + b = -c ; b + c = -a ; c + a = -b
A = (1+a/b)(1+b/c)(1+c/a)
= a+b/b . b+c/c . c+a/a
= -c/b . -a/c . -b/a
= -abc/abc
= -1
TH 2 : a^2 + b^2 + c^2 - ab - bc - ac = 0
<=> 2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ac = 0
<=> (a-b)^2 + (b-c)^2 + (a-c)^2 = 0 ( làm tắt
)
Do (a-b)^2 >= 0 ; (b-c)^2 >= 0 ; (a-c)^2 >= 0
=> a - b = 0 ; b - c = 0 ; a - c = 0
=> a = b = c
A = (1+a/b) ( 1+b/c) ( 1 + c/a)
= (1+1)(1+1)(1+1)
= 2 . 2 . 2
= 8
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