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

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

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

Ta có :

x + y + z = 10

=> y + z = 10 - x

\(Q=x^2-\left(10-x\right)^2-20x\)

\(Q=x^2-\left(100-20x+x^2\right)-20x\)

\(Q=x^2-100+20x-x^2-20x\)

\(Q=-100\)

Do x+y+z=10 nên y+z=10-x, ta có:

  \(x^2-20x-y^2-2yz-z^2\) nên bằng \(x^2-20x\left(y^2+2yz+z^2\right)\)

\(=x^2-20x-\left(y+z\right)^2\) <=> = \(x^2-20x-\left(10-x\right)^2\)

=\(x^2-20x-100+20x-x^2\)

và bằng -100 .... tck nha bạn

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)

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