Lời giải:

$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}$

$\Rightarrow (\frac{1}{x}+\frac{1}{y})+(\frac{1}{z}-\frac{1}{x+y+z})=0$

$\Leftrightarrow \frac{x+y}{xy}+\frac{x+y}{z(x+y+z)}=0$

$\Leftrightarrow (x+y)(\frac{1}{xy}+\frac{1}{z(x+y+z)})=0$

$\Leftrightarrow (x+y).\frac{z(x+y+z)+xy}{xyz(x+y+z)}=0$

$\Leftrightarrow (x+y).\frac{(z+x)(z+y)}{xyz(x+y+z)}=0$

$\Leftrightarrow (x+y)(y+z)(x+z)=0$

$\Leftrightarrow x=-y$ hoặc $y=-z$ hoặc $z=-x$

Nếu $x=-y$ thì:

$P=\frac{3}{4}+[(-y)^8-y^8](y^9+z^9)(z^{10}-x^{10})=\frac{3}{4}+0.(y^9+z^9)(z^{10}-x^{10})=\frac{3}{4}$

Nếu $y=-z$ thì:

$P=\frac{3}{4}+(x^8-y^8)[(-z)^9+z^9](z^{10}-x^{10})=\frac{3}{4}+(x^8-y^8).0.(z^{10}-x^{10})=\frac{3}{4}$

Nếu $z=-x$ thì:

$P=\frac{3}{4}+(x^8-y^8)(y^9+z^9)[(-x)^{10}-x^{10}]=\frac{3}{4}+(x^8-y^8)(y^9+z^9).0=\frac{3}{4}$

Từ \(x\left(\dfrac{1}{y}+\dfrac{1}{z}\right)+y\left(\dfrac{1}{z}+\dfrac{1}{x}\right)+z\left(\dfrac{1}{x}+\dfrac{1}{y}\right)=-2\) ta có:

\(x^2y+y^2z+z^2x+xy^2+yz^2+zx^2+2xyz=0\)

\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+y=0\\y+z=0\\z+x=0\end{matrix}\right.\).

Không mất tính tổng quát, giả sử x + y = 0

\(\Leftrightarrow x=-y\)

\(\Leftrightarrow x^3=-y^3\).

Kết hợp với \(x^3+y^3+z^3=1\) ta có \(z^3=1\Leftrightarrow z=1\).

Vậy \(P=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{-y}+\dfrac{1}{y}+\dfrac{1}{1}=1\).

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

(y + z - x)/x = (z + x - y)/y = (x + y - z)/z = 1

--> y + z - x = x; z + x - y = y; x + y - z = z

--> y + z = 2x; z + x = 2y; x + y = 2z

Ta có: 

B = (x + y)/y.(y + z)/z.(z + x)/x

= 2z/y.2x/z.2y/x = 8

Ta có:

\(x^2+1=x^2+xy+yz+zx\)

           \(=x\left(x+y\right)+z\left(x+y\right)=\left(x+y\right)\left(x+z\right)\)

Tương tự:

\(\left\{{}\begin{matrix}y^2+1=\left(y+z\right)\left(y+x\right)\\z^2+1=\left(z+y\right)\left(z+x\right)\end{matrix}\right.\)

\(A=x\sqrt{\dfrac{\left(x+y\right)\left(y+z\right)\left(z+x\right)\left(y+z\right)}{\left(x+y\right)\left(z+x\right)}}+y\sqrt{\dfrac{\left(z+x\right)\left(y+z\right)\left(x+y\right)\left(z+x\right)}{\left(x+y\right)\left(y+z\right)}}+z\sqrt{\dfrac{\left(x+y\right)\left(z+x\right)\left(y+z\right)\left(x+y\right)}{\left(z+x\right)\left(y+z\right)}}\)

\(=x\left|y+z\right|+y\left|z+x\right|+z\left|x+y\right|\)

TH1: x,y,z <0

\(A=-x\left(y+z\right)-y\left(z+x\right)-z\left(x+y\right)=-2\)

TH2: x,y,z>0

\(A=x\left(y+z\right)+y\left(z+x\right)+z\left(x+y\right)=2\)

Ta có \(1+z^2=xy+yz+zx+z^2\)

\(=y\left(x+z\right)+z\left(x+z\right)\)

\(=\left(x+z\right)\left(y+z\right)\)

CMTT, \(1+x^2=\left(x+y\right)\left(x+z\right)\) và \(1+y^2=\left(x+y\right)\left(y+z\right)\)

Do đó \(\sqrt{\dfrac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}\) \(=\sqrt{\dfrac{\left(x+y\right)\left(y+z\right)\left(x+z\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}\)

\(=\sqrt{\left(y+z\right)^2}\) \(=\left|y+z\right|\)

 Tương tự như thế, ta được

\(A=x\left|y+z\right|+y\left|z+x\right|+z\left|x+y\right|\)

 Cái này không tính ra số cụ thể được nhé bạn. Nó còn phải tùy vào dấu của \(x+y,y+z,z+x\) nữa.

Hướng dẫn: đặt \(A=\dfrac{y^4}{\left(x^2+y^2\right)\left(x+y\right)}+\dfrac{z^4}{\left(y^2+z^2\right)\left(y+z\right)}+\dfrac{x^4}{\left(z^2+x^2\right)\left(z+x\right)}\)

Khi đó \(F-A=x-y+y-z+z-x=0\Rightarrow F=A\)

\(\Rightarrow2F=F+A=\sum\dfrac{x^4+y^4}{\left(x^2+y^2\right)\left(x+y\right)}\ge\sum\dfrac{\left(x^2+y^2\right)^2}{2\left(x^2+y^2\right)\left(x+y\right)}\ge\sum\dfrac{\left(x+y\right)^2\left(x^2+y^2\right)}{4\left(x^2+y^2\right)\left(x+y\right)}\)

\(\Rightarrow2F\ge\dfrac{x+y+z}{2}\Rightarrow F\ge\dfrac{x+y+z}{4}\)

\(P\le\sqrt{3\left(\sum\dfrac{1}{\left(x+y\right)^2+\left(x+1\right)^2+4}\right)}\le\sqrt{3\left(\sum\dfrac{1}{4xy+4x+4}\right)}\)

\(P\le\sqrt{\dfrac{3}{4}\sum\left(\dfrac{1}{xy+x+1}\right)}=\dfrac{\sqrt{3}}{2}\)

\(P_{max}=\dfrac{\sqrt{3}}{2}\) khi \(x=y=z=1\)