Toán Lớp 8: Cho `a+b+c=1` và `a^3+b^3+c^3=3abc`. Chứng tỏ `a=b=c`.

Question

Toán Lớp 8:  Cho a+b+c=1 và a^3+b^3+c^3=3abc. Chứng tỏ a=b=c.

hoquyly · Answer

$  $   a^3 + b^3 +c^3=3abc   &rArr; a^3 + b^3 +c^3-3abc=0   &rArr; (a+b)^3 +c^3-  3ab (a+b) -3abc=0   &rArr; (a+b+c)[(a+b)^2 - (a+b)c +c^2] -3ab (a+b+c)=0   &rArr; (a+b+c)(a^2 + b^2 +c^2 - ab-bc-ac)=0   &rArr; a^2 +b^2+c^2 -ab-ac-bc=0 (V&igrave; a+b+c=1)   &rArr; 2a^2 +2b^2 +2c^2-2ab-2ac-2bc=0   &rArr; (a-b)^2 +(b-c)^2 + (a-c)^2=0   V&igrave; (a-b)^2 &ge;0, (b-c)^2 &ge;0, (a-c)^2 &ge;0&forall;a,b,c   => (a-b)^2 +(b-c)^2 + (a-c)^2 &ge;0&forall;a,b,c   Dấu '=' xảy ra khi :   (a-b)^2=0 ,(b-c)^2=0, (a-c)^2=0   &hArr; a-b=0,b-c=0,a-c=0   &hArr;a=b,b=c,a=c   &hArr;a=b=c

myngocla · Answer

+)a^3+b^3+c^3=3abc
<=>a^3+b^3+c^3-3abc=0
<=>a^3+3a^2b+3ab^2+b^3+c^3-3a^2b-3ab^2-3abc=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^2+2ab+b^2-ca-bc+c^2-3ab)=0
<=>(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0
Vì a+b+c=1
=>a^2+b^2+c^2-ab-bc-ca=0
<=>2(a^2+b^2+c^2-ab-bc-ca)=0
<=>2a^2+2b^2+2c^2-2ab-2bc-2ca=0
<=>(a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ca+a^2)=0
<=>(a-b)^2+(b-c)^2+(c-a)^2=0
<=>{(a-b=0),(b-c=0),(c-a=0):}
<=>{(a=b),(b=c),(c=a):}
->a=b=c
Vậy : a=b=c