\(\frac{xy}{x+y}=\frac{yz}{y+z}=\frac{zx}{z+x}\Rightarrow\frac{xyz}{z\left(x+y\right)}=\frac{xyz}{x\left(y+z\right)}=\frac{xyz}{y\left(z+x\right)}\)

 \(\frac{xyz}{z\left(x+y\right)}=\frac{xyz}{x\left(y+z\right)}\Rightarrow z\left(x+y\right)=x\left(y+z\right)\Rightarrow xz+yz=xy+xz\Rightarrow yz=xy\Rightarrow z=x\)

CM tương tự ta cũng có : \(x=y;y=z\)

\(\Rightarrow x=y=z\) Thay vào B ta được :

\(B=\frac{x^3+y^3+z^3}{x^2y+y^2z+z^2x}=\frac{x^3+x^3+x^3}{x^2x+x^2x+x^2x}=\frac{3x^3}{3x^3}=1\)

a) \(xy+x+2y=5\\ \Rightarrow y\left(x+2\right)+x+2=5+2\\ \Rightarrow\left(x+2\right)\left(y+1\right)=7\)

Ta xét bảng:

Vậy \(\left(x;y\right)\in\left\{\left(-1;6\right);\left(5;0\right);\left(-3;-8\right);\left(-9;-2\right)\right\}\)

b) \(xy-3x-y=0\\ \Rightarrow x\left(y-3\right)-y+3=3\\ \Rightarrow\left(y-3\right)\left(x-1\right)=3\)

Ta xét bảng:

Vậy \(\left(x;y\right)\in\left\{\left(2;6\right);\left(4;4\right);\left(0;0\right);\left(-2;2\right)\right\}\)

c) \(xy+2x+2y=-16\\ \Rightarrow x\left(y+2\right)+2y+4=-12\\ \Rightarrow\left(y+2\right)\left(x+2\right)=-12\)

Ta xét bảng:

Vậy \(\left(x;y\right)\in\left\{\left(-1;-14\right);\left(0;-8\right);\left(1;-6\right);\left(2;-5\right);\left(4;-4\right);\left(10;-3\right);\left(-3;10\right);\left(-4;4\right);\left(-5;2\right);\left(-6;1\right);\left(-8;0\right);\left(-14;-1\right)\right\}\)