12. [tex]a^3+b^3+c^3-3abc=(a+b)^3-3ab(a+b)+c^3-3abc=(a+b)^3+c^3-3ab(a+b+c)=(a+b+c)[(a+b)^2-c(a+b)+c^2]-3(a+b+c).ab=(a+b+c)[(a+b)^2-c(a+b)+c^2-3ab]=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)[/tex]
13. [tex]\frac{1}{1+y+yz}=\frac{x}{x+xy+xyz}=\frac{x}{1+x+xy};\frac{1}{1+z+zx}=\frac{xy}{xy+xyz+x.xyz}=\frac{xy}{1+x+xy}[/tex]
14. Ta có: [tex]a^2+b^2+c^2=1\Rightarrow -1\leq a,b,c\leq 1\Rightarrow a^2\geq a^3,b^2\geq b^3,c^2\geq c^3\Rightarrow a^2+b^2+c^2\geq a^3+b^3+c^3[/tex]
Dấu "=" xảy ra khi a,b,c bằng 0 hoặc 1. Mà [TEX]a^3+b^3+c^3=1[/TEX] nên (a,b,c)=(1,0,0) và các hoán vị.
Từ đó S = 1.
15. [tex]a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=(a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2]=0\Leftrightarrow a+b+c=0 [/tex] hoặc [TEX]a=b=c[/TEX]
Khi a+b+c=0 thì [tex]a=-(b+c)\Rightarrow a^2=b^2+c^2+2bc\Rightarrow a^2-b^2-c^2=2bc[/tex]
16. Cộng vế theo vế ta có: [tex]a^3+b^3-3(a^2+b^2)+5(a+b)-6=0\Rightarrow (a+b)(a^2-ab+b^2)-2(a^2+b^2-ab)-(a^2+b^2+2ab)+2(a+b)+3(a+b)-6=0\Rightarrow (a+b-2)(a^2-ab+b^2)-(a+b)^2+2(a+b)+3(a+b-2)=0\Rightarrow (a+b-2)(a^2-ab+b^2)-(a+b)(a+b-2)+3(a+b-2)=0\Rightarrow (a+b-2)(a^2-ab+b^2-a-b+3)=0\Rightarrow (a+b-2)[(a-\frac{1}{2}b-\frac{1}{2})^2+\frac{3}{4}b^2-\frac{3}{2}b+\frac{11}{4}]=0\Rightarrow a+b=2[/tex]
17. Tương tự bài 13.
18. [tex][\frac{1}{(a-b)^2}+\frac{1}{(b-c)^2}+\frac{1}{(c-a)^2}]^2=\frac{1}{(a-b)^2}+\frac{1}{(b-c)^2}+\frac{1}{(c-a)^2}+\frac{2}{(a-b)(b-c)}+\frac{2}{(b-c)(c-a)}+\frac{2}{(c-a)(a-b)}=\frac{1}{(a-b)^2}+\frac{1}{(b-c)^2}+\frac{1}{(c-a)^2}+\frac{2(a-b+b-c+c-a)}{(a-b)(b-c)(c-a)}=\frac{1}{(a-b)^2}+\frac{1}{(b-c)^2}+\frac{1}{(c-a)^2}[/tex]
19. [tex]\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Rightarrow \frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}\Rightarrow A=xyz(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3})=xyz.\frac{3}{xyz}=3[/tex]
20. [tex]\frac{1}{x+y+z}=\frac{xy+yz+zx}{xyz}\Rightarrow (x+y+z)(xy+yz+zx)-xyz=0\Rightarrow (x+y)(y+z)(z+x)=0[/tex]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
