Lời giải:

Từ \(a+b+c+ab+bc+ac=0\)

\(\Rightarrow a+b+c+ab+bc+ac+abc+1=1\)

\(\Leftrightarrow (a+1)(b+1)(c+1)=1\)

Đặt \(\left\{\begin{matrix} a+1=x\\ b+1=y\\ c+1=z\end{matrix}\right.\Rightarrow xyz=1\)

Biểu thức trở thành:

\(A=\frac{1}{(a+2)+a+b+ab+1}+\frac{1}{(b+2)+b+c+bc+1}+\frac{1}{(c+2)+c+a+ac+1}\)

\(A=\frac{1}{(a+2)+(a+1)(b+1)}+\frac{1}{(b+2)+(b+1)(c+1)}+\frac{1}{(c+2)+(c+1)(a+1)}\)

\(A=\frac{1}{x+1+xy}+\frac{1}{y+1+yz}+\frac{1}{z+1+zx}\)

\(A=\frac{z}{xz+z+xyz}+\frac{zx}{yxz+xz+yz.xz}+\frac{1}{z+1+xz}\)

hay \(A=\frac{z}{xz+z+1}+\frac{xz}{1+xz+z}+\frac{1}{z+1+xz}\) (thay \(xyz=1\))

\(\Leftrightarrow A=\frac{z+xz+1}{xz+z+1}=1\)

Vậy \(A=1\)

Lời giải:

\(a+b+c+ab+bc+ac+abc=0\)

\(\Leftrightarrow (a+b+ab+1)+c+bc+ac+abc=1\)

\(\Leftrightarrow (a+b+ab+1)+c(1+b+a+ab)=1\)

\(\Leftrightarrow (a+1)(b+1)+c(a+1)(b+1)=1\)

\(\Leftrightarrow (a+1)(b+1)(c+1)=1\)

Đặt \((a+1,b+1,c+1)=(x,y,z)\Rightarrow (a,b,c)=(x-1,y-1,z-1)\) và \(xyz=1\)

Khi đó:

\(P=\frac{1}{3+2(x-1)+y-1+(x-1)(y-1)}+\frac{1}{3+2(y-1)+z-1+(y-1)(z-1)}+\frac{1}{3+2(z-1)+x-1+(x-1)(z-1)}\)

\(=\frac{1}{x+xy+1}+\frac{1}{y+yz+1}+\frac{1}{z+xz+1}\)

\(=\frac{yz}{xyz+xy.yz+yz}+\frac{1}{y+yz+1}+\frac{y}{zy+xz.y+y}\)

\(=\frac{yz}{1+y+yz}+\frac{1}{y+yz+1}+\frac{y}{yz+1+y}=\frac{yz+1+y}{yz+1+y}=1\)

Ta có đpcm.

1/\(=4a^2+4b^2+c^2+8ab-4bc-4ca+4b^2+4c^2+a^2+8bc-4ca-4ab+4a^2+4c^2+b^2+8ca-4bc-4ab=\)

\(=9a^2+9b^2+9c^2=9\left(a^2+b^2+c^2\right)\)

2/

Ta có

\(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)\ge0\)

\(\Leftrightarrow a^2+b^2+c^2\ge-2\left(ab+bc+ca\right)=2\)

\(\Rightarrow P=9\left(a^2+b^2+c^2\right)\ge18\)

\(\Rightarrow P_{min}=18\)