Ta có:\(10=2xyz\)

=> \(P=\frac{1}{2x+2xz+1}+\frac{2xy}{y+2xy+10}+\frac{10z}{10z+yz+10}\) 

        \(=\frac{1}{2x+2xz+1}+\frac{2xy}{y+2xy+2xyz}+\frac{2xyz^2}{2xyz^2+yz+2xyz}\)

          \(=\frac{1}{2x+2xz+1}+\frac{2x}{1+2x+2xz}+\frac{2xz}{2xz+1+2x}\)

          \(=1\)

Vậy P=1

Ta có \(\frac{x+2xy+1}{x+xy+xz+1}=\frac{x+2xy+xyz}{x+xy+xz+xyz}=\frac{1+2y+yz}{\left(y+1\right)\left(z+1\right)}\)

Tương tự => \(M=\frac{1+2y+yz}{\left(y+1\right)\left(z+1\right)}+\frac{1+2z+zx}{\left(1+x\right)\left(z+1\right)}+\frac{1+2x+xy}{\left(1+x\right)\left(y+1\right)}\)

=> \(M=\frac{\left(1+2y+yz\right)\left(1+x\right)+\left(1+2z+zx\right)\left(1+y\right)+\left(1+2x+xy\right)\left(1+z\right)}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\)

=>\(M=\frac{6+3\left(x+y+z\right)+3\left(xy+yz+xz\right)}{2+\left(x+y+z\right)+\left(xy+yz+xz\right)}=3\)

Hướng dẫn :\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Rightarrow\frac{xy+yz+zx}{xyz}=0\Rightarrow xy+yz+zx=0\)

Thay vào:\(x^2+2yz=x^2+yz+yz=x^2+yz-xy-zx=x\left(x-y\right)-z\left(x-y\right)=\left(x-y\right)\left(x-z\right)\)

Tương tự thay vào mà quy đồng

Áp dụng bất đẳng thức Cauchy , ta có : 

\(x+y+z\ge3\sqrt[3]{xyz}\)

<=> \(xyz\ge3\sqrt[3]{xyz}\)

<=> \(x^3y^3z^3\ge27xyz\)

<=> \(x^2y^2z^2\ge27\)

<=> \(\sqrt[3]{x^2y^2z^2}\ge3\)

Ta có 

\(P=\frac{1}{x^2+yz+yz}+\frac{1}{y^2+zx+zx}+\frac{1}{z^2+xy+xy}\le\frac{1}{3\sqrt[3]{x^2y^2z^2}}+\frac{1}{3\sqrt[3]{x^2y^2z^2}}+\frac{1}{3\sqrt[3]{x^2y^2z^2}}\)

                                                                                                                  \(=\frac{1}{\sqrt[3]{x^2y^2z^2}}\le\frac{1}{3}\)

Vậy Max = 1/3 

Sửa lại đề : tính \(A=\frac{yz}{x^2+2yz}+\frac{xz}{y^2+2xz}+\frac{xy}{z^2+2xy}\)

Từ \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Leftrightarrow\frac{xy+yz+xz}{xyz}=0\Rightarrow xy+yz+xz=0\)

\(\Rightarrow yz=-xy-xz\)

\(\Rightarrow x^2+2yz=x^2+yz-xy-xz=x\left(x-y\right)-z\left(x-y\right)=\left(x-z\right)\left(x-y\right)\)

CM tương tự ta cx có : \(\hept{\begin{cases}y^2+2xz=\left(y-x\right)\left(y-z\right)\\z^2+2xy=\left(z-x\right)\left(z-y\right)\end{cases}}\)

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

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

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

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

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

\(=\frac{\left(y-z\right)\left(y-x\right)z+\left(x-y\right)\left(y-z\right)x}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)

\(=\frac{\left(y-z\right)\left(x-y\right)\left(x-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=1\)