\(Q=x^2-y^2-z^2-2yz-20x\)

=> \(Q=x^2-y^2-z^2-2yz-20x+100-100\)

=> \(Q=\left(x^2-20x+100\right)-\left(y^2+2yz+z^2\right)-100\)

=> \(Q=\left(x-10\right)^2-\left(y+z\right)^2-10^2\)

=> \(Q=\left(x-10-y-z\right)\left(x-10+y+z\right)-10^2\)

=> \(Q=\left[x-\left(x+y+z\right)-y-z\right]\left[x-\left(x+y+z\right)+y+z\right]-10^2\)

=> \(Q=\left[x-\left(x+y+z\right)-y-z\right]\left[x-\left(x+y+z\right)+y+z\right]-10^2\)

=> \(Q=(-2y-2z).0-10^2\)

=> \(Q=0-10^2\)

=> \(Q=-100\)

TA có :

\(x+y+z=10\Rightarrow\left(x+y+z\right)^2=100\Leftrightarrow x^2+y^2+z^2+2xy+2yz+2xz\)\(Q=x^2-y^2-z^2-2yz-20x\)

\(\Rightarrow100+Q=x^2+y^2+z^2+2xy+2yz+2xz+x^2-y^2-z^2-2yz-20x\)\(=2x^2+2xy+2xz-20x\)

\(=2x\left(x+y+z-10\right)\)

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

\(\Rightarrow100+Q=0\Rightarrow Q=-100\)

Bài này ez thôi, làm mãi rồi.

Theo đề bài, ta có: \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)

=>\(\dfrac{xy+yz+xz}{xyz}=0\)

=> xy+yz+zx=0

=> \(\left\{{}\begin{matrix}xy=-yz-zx\\yz=-xy-zx\\zx=-xy-yz\end{matrix}\right.\)

Ta có: x²+2yz=x²+yz-xy-zx=(x-y)(x-z)

           y²+2xz=y²+xz-xy-yz=(x-y)(z-y)

           z²+2xy=z²+xy-yz-xz=(x-z)(y-z)

=> \(\dfrac{yz}{\left(x-y\right)\left(x-z\right)}+\dfrac{xz}{\left(x-y\right)\left(z-y\right)}+\dfrac{xy}{\left(x-z\right)\left(y-z\right)}=\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\dfrac{\left(x-y\right)\left(x-z\right)\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=1\)