Question

cho x ,y ,z thỏa mãn xy+xz+yz =0 và x+y+z= -1

Tính \(M=\dfrac{xy}{z}+\dfrac{xz}{y}+\dfrac{yz}{x}\)

Answer 1

Để M xác định thì \(x,y,z\ne0\)

\(xy+xz+yz=0\Rightarrow\left\{{}\begin{matrix}\dfrac{xy}{z}+x+y=0\\\dfrac{xz}{y}+x+z=0\\\dfrac{yz}{x}+y+z=0\end{matrix}\right.\)

Cộng vế với vế ta được:

\(\dfrac{xy}{z}+\dfrac{xz}{y}+\dfrac{yz}{x}+2\left(x+y+z\right)=0\)

\(\Leftrightarrow M+2.\left(-1\right)=0\Rightarrow M=2\)

Answer 2

Ta có :

\(xy+yz+xz=0\\ \Rightarrow\left[{}\begin{matrix}xy=-xz-yz=-z\left(x+y\right)\\yz=-xy-xz=-x\left(y+z\right)\\xz=-xy-yz=-y\left(x+z\right)\end{matrix}\right.\)

\(M=\dfrac{xy}{z}+\dfrac{xz}{y}+\dfrac{yz}{x}=\dfrac{-z\left(x+y\right)}{z}+\dfrac{-y\left(x+z\right)}{y}+\dfrac{-x\left(y+z\right)}{x}\\ =-\left(x+y\right)-\left(x+z\right)-\left(y+z\right)=-x-y-x-z-y-z\\ =-2\left(x+y+z\right)=\left(-2\right)\cdot\left(-1\right)=2\)

\(\Rightarrow M=2\)