Đáp án: $\dfrac{33}2$

Giải thích các bước giải:

Ta có:

$x^2-3x-1$

$=x^2-4x+x-1$

$=x^2-(x+y+z)x+x-1$

$=-xy-xz+x-1$

$=-xy-xz+x+xyz$

$=(xyz-xy)-(xz-x)$

$=xy(z-1)-x(z-1)$

$=x(y-1)(z-1)$

$\to \dfrac{x}{x^2-3x-1}=\dfrac{x}{x(y-1)(z-1)}=\dfrac1{(y-1)(z-1)}$

Tương tự:

$\dfrac{y}{y^2-3y-1}=\dfrac1{(z-1)(x-1)}$

$\dfrac{z}{z^2-3z-1}=\dfrac1{(x-1)(y-1)}$

$\to \dfrac1{(x-1)(y-1)}+\dfrac1{(y-1)(z-1)}+\dfrac1{(z-1)(x-1)}=\dfrac49$

$\to (x-1)+(y-1)+(z-1)=\dfrac49(x-1)(y-1)(z-1)$

$\to x+y+z-3=\dfrac49(x-1)(y-1)(z-1)$

$\to4-3=\dfrac49(x-1)(y-1)(z-1)$

$\to 1=\dfrac49(x-1)(y-1)(z-1)$

$\to (x-1)(y-1)(z-1)=\dfrac94$

$\to xyz-(xy+yz+zx)+(x+y+z)-1=\dfrac94$

$\to (-1)-(xy+yz+zx)+4-1=\dfrac94$

$\to xy+yz+zx=-\dfrac14$

Ta có:

$P=x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+zx)=4^2-2\cdot (-\dfrac14)=\dfrac{33}2$