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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
\(a^2-2b+6b+b^2=-10\)
\(\Leftrightarrow a^2-2a+6b+b^2+10=0\)
\(\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2+6b+9\right)=0\)
\(\Leftrightarrow\left(a-1\right)^2+\left(b+3\right)^2=0\left(1\right)\)
Vì \(\hept{\begin{cases}\left(a-1\right)^2\ge0\forall a\\\left(b+3\right)^2\ge0\forall b\end{cases}\Leftrightarrow\hept{\begin{cases}\left(a-1\right)^2=0\\\left(b+3\right)^2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}a=1\\b=-3\end{cases}}}\)
\(L=\frac{x+y}{z}+1+\frac{y+z}{x}+1+\frac{x+z}{y}+1-3\)
\(=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)-3=0-3=-3\)
\(\frac{x+y}{z}+\frac{y+z}{x}+\frac{z+x}{y}=\frac{x+y+z}{z}-1+\frac{x+y+z}{y}-1+\frac{x+y+z}{x}-1\)
\(=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)-3=0-3=-3\)
\(A=\frac{x+y}{z}+\frac{x+z}{y}+\frac{y+z}{x}\)
\(=\frac{x+y}{z}+1+\frac{y+z}{x}+1+\frac{x+z}{y}+1-3\)
\(=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)-3\)
\(=0-3\)
\(=-3\)
thay z = -(x+y) , y = -(z+x),... vao
=> Duoc bieu thuc trong do co 1/xy + 1/yz + 1/zx = (x+y+z)/xyz = 0
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 : \(A=x.\left(\frac{-1}{x}\right)+y.\left(-\frac{1}{y}\right)+z.\left(-\frac{1}{z}\right)=\left(-1\right)+\left(-1\right)+\left(-1\right)=-3\)