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* Nếu x + y + z = 0
\(A=\left(1+\dfrac{y}{x}\right)\left(1+\dfrac{z}{y}\right)\left(1+\dfrac{x}{z}\right)\)
\(=\dfrac{x+y}{x}\cdot\dfrac{y+z}{y}\cdot\dfrac{z+x}{z}=\dfrac{\left(-z\right)}{x}\cdot\dfrac{\left(-x\right)}{y}\cdot\dfrac{\left(-y\right)}{z}=\dfrac{\left(-x\right)\left(-y\right)\left(-z\right)}{xyz}=-\dfrac{xyz}{xyz}=-1\)
* Nếu x + y + z khác 0
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x-y-z}{x}=\dfrac{y-x-z}{y}=\dfrac{-x-y+z}{z}=\dfrac{x-y-z+y-x-z-x-y+z}{x+y+z}=\dfrac{-x-y-z}{x+y+z}=-1\)
\(\Rightarrow\left\{{}\begin{matrix}x-y-z=-x\\y-x-z=-y\\-x-y+z=-z\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y+z=2x\\z+x=2y\\x+y=2z\end{matrix}\right.\Rightarrow x=y=z\)
\(\Rightarrow A=\left(1+\dfrac{y}{x}\right)\left(1+\dfrac{z}{y}\right)\left(1+\dfrac{x}{z}\right)\)
\(=\left(1+1\right)\left(1+1\right)\left(1+1\right)=2\cdot2\cdot2=8\)
Ta có: \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Rightarrow\dfrac{xy+yz+xz}{xyz}=0\Rightarrow xy+yz+xz=0\)
\(\Rightarrow xy=-yz-xz;yz=-xy-xz;xz=-xy-yz\)
Ta lại có: \(A=\dfrac{x+y}{z}+\dfrac{x+z}{y}+\dfrac{y+z}{x}=\dfrac{x^2+xy}{xz}+\dfrac{z^2+xz}{yz}+\dfrac{y^2+yz}{xy}\)
\(=\dfrac{x^2-yz-xz}{xz}+\dfrac{z^2-xy-yz}{yz}+\dfrac{y^2-xy-xz}{xy}\)
\(=\dfrac{x\left(x-z\right)}{xz}-\dfrac{yz}{xz}+\dfrac{z\left(z-y\right)}{yz}-\dfrac{xy}{yz}+\dfrac{y\left(y-x\right)}{xy}-\dfrac{xz}{xy}\)
\(=\dfrac{x-z}{z}-\dfrac{y}{x}+\dfrac{z-y}{y}-\dfrac{x}{z}+\dfrac{y-x}{x}-\dfrac{z}{y}\)
\(=\dfrac{x-z-x}{z}+\dfrac{z-y-z}{y}+\dfrac{y-x-y}{x}=\dfrac{-z}{z}+\dfrac{-y}{y}+\dfrac{-x}{x}\)
\(=-1-1-1=-3\). Vậy A=-3
\(A=\dfrac{x+y}{z}+\dfrac{y+z}{x}+\dfrac{z+x}{y}\) (đã sửa đề)
\(A+3=\dfrac{x+y+z}{z}+\dfrac{x+y+z}{x}+\dfrac{x+y+z}{y}\)
\(A+3=\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=0\)
\(A=-3\)
thank you