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\(Q=x^2-y^2-z^2-2yz-20x\)
=> \(Q=x^2-y^2-z^2-2yz-20x+100-100\)
=> \(Q=\left(x^2-20x+100\right)-\left(y^2+2yz+z^2\right)-100\)
=> \(Q=\left(x-10\right)^2-\left(y+z\right)^2-10^2\)
=> \(Q=\left(x-10-y-z\right)\left(x-10+y+z\right)-10^2\)
=> \(Q=\left[x-\left(x+y+z\right)-y-z\right]\left[x-\left(x+y+z\right)+y+z\right]-10^2\)
=> \(Q=\left[x-\left(x+y+z\right)-y-z\right]\left[x-\left(x+y+z\right)+y+z\right]-10^2\)
=> \(Q=(-2y-2z).0-10^2\)
=> \(Q=0-10^2\)
=> \(Q=-100\)
TA có :
\(x+y+z=10\Rightarrow\left(x+y+z\right)^2=100\Leftrightarrow x^2+y^2+z^2+2xy+2yz+2xz\)\(Q=x^2-y^2-z^2-2yz-20x\)
\(\Rightarrow100+Q=x^2+y^2+z^2+2xy+2yz+2xz+x^2-y^2-z^2-2yz-20x\)\(=2x^2+2xy+2xz-20x\)
\(=2x\left(x+y+z-10\right)\)
\(=2x\left(x+y+z-x-y-z\right)\)
\(\Rightarrow100+Q=0\Rightarrow Q=-100\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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)
=> \(\dfrac{yz}{\left(x-y\right)\left(x-z\right)}+\dfrac{xz}{\left(x-y\right)\left(z-y\right)}+\dfrac{xy}{\left(x-z\right)\left(y-z\right)}=\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\dfrac{\left(x-y\right)\left(x-z\right)\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(Q=x^2-y^2-z^2-2yz-20x\)
\(Q=x^2-\left(y^2+z^2+2yz\right)-20x\)
\(Q=x^2-\left(y+z\right)^2-20x\)
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