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Ta có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Leftrightarrow\hept{\begin{cases}\frac{1}{x}+\frac{1}{y}=-\frac{1}{z}\\\frac{1}{y}+\frac{1}{z}=-\frac{1}{x}\\\frac{1}{x}+\frac{1}{z}=-\frac{1}{y}\end{cases}}\) (*)
Ta có: \(A=\frac{x+y}{z}+\frac{x+z}{y}+\frac{y+z}{x}\)
\(=\frac{x}{z}+\frac{y}{z}+\frac{x}{y}+\frac{x}{y}+\frac{y}{x}+\frac{z}{x}\)
\(=\left(\frac{x}{z}+\frac{x}{y}\right)+\left(\frac{y}{x}+\frac{y}{z}\right)+\left(\frac{z}{x}+\frac{z}{y}\right)\)
\(=x\left(\frac{1}{z}+\frac{1}{y}\right)+y\left(\frac{1}{x}+\frac{1}{z}\right)+z\left(\frac{1}{x}+\frac{1}{y}\right)\)
Thay (*) vào,ta có : \(A=x.\left(\frac{-1}{x}\right)+y.\left(-\frac{1}{y}\right)+z.\left(-\frac{1}{z}\right)=\left(-1\right)+\left(-1\right)+\left(-1\right)=-3\)
\(\frac{x+y}{z}+\frac{y+z}{x}+\frac{z+x}{y}=\frac{x+y+z}{z}-1+\frac{x+y+z}{y}-1+\frac{x+y+z}{x}-1\)
\(=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)-3=0-3=-3\)
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)=> (x+y+z)\(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)=0
=> \(\frac{y+z}{x}+\frac{x+z}{y}+\frac{x+y}{z}+3=0\)
=> \(\frac{y+z}{x}+\frac{x+z}{y}+\frac{x+y}{z}=-3\)
\(A=\frac{x+y}{z}+\frac{x+z}{y}+\frac{y+z}{x}\)
\(=\frac{x+y}{z}+1+\frac{y+z}{x}+1+\frac{x+z}{y}+1-3\)
\(=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)-3\)
\(=0-3\)
\(=-3\)
Ta có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\Rightarrow\frac{1}{x}+\frac{1}{y}=-\frac{1}{z}\)
\(\frac{1}{x}+\frac{1}{z}=-\frac{1}{y}\)
\(\frac{1}{y}+\frac{1}{z}=-\frac{1}{x}\)
\(A=\frac{x+y}{z}+\frac{x+z}{y}+\frac{y+z}{x}=\frac{x}{z}+\frac{y}{z}+\frac{x}{y}+\frac{z}{y}+\frac{y}{x}+\frac{z}{x}\)
\(=\left(\frac{y}{z}+\frac{y}{x}\right)+\left(\frac{x}{z}+\frac{x}{y}\right)+\left(\frac{z}{y}+\frac{z}{x}\right)\)
\(=y\left(\frac{1}{z}+\frac{1}{x}\right)+x\left(\frac{1}{z}+\frac{1}{y}\right)+z\left(\frac{1}{y}+\frac{1}{x}\right)\)
\(=y.\frac{-1}{y}+x.\frac{-1}{x}+z.\frac{-1}{z}=-1-1-1=-3\)
Vậy nên A = -3
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\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Leftrightarrow\frac{yz}{xyz}+\frac{xz}{xyz}+\frac{xy}{xyz}=0\)
\(\Leftrightarrow\frac{yz+xz+xy}{xyz}=0\Leftrightarrow yz+xz+xy=0\)
\(A=\frac{x+y}{z}+\frac{x+z}{y}+\frac{y+z}{x}=\frac{xy\left(x+y\right)}{xyz}+\frac{xz\left(x+z\right)}{xyz}+\frac{yz\left(y+z\right)}{xyz}\)
\(=\frac{x^2y+xy^2}{xyz}+\frac{x^2z+xz^2}{xyz}+\frac{y^2z+yz^2}{xyz}=\frac{x^2y+xy^2+x^2z+xz^2+y^2z+yz^2}{xyz}\)
\(=\frac{\left(x^2y+x^2z+xyz\right)+\left(xy^2+y^2z+xyz\right)+\left(xz^2+yz^2+xyz\right)-3xyz}{xyz}\)
\(=\frac{x\left(xy+xz+yz\right)+y\left(xy+yz+xz\right)+z\left(xz+yz+xy\right)-3xyz}{xyz}\)
\(=\frac{\left(x+y+z\right)\left(xz+yz+xy\right)-3xyz}{xyz}=\frac{\left(x+y+z\right).0-3xyz}{xyz}=\frac{-3xyz}{xyz}-3\)