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tk cho mk nhé
Do \(xyz\ne0\) ta có:
\(\dfrac{1}{xy}+\dfrac{1}{xz}+\dfrac{1}{yz}=0\Leftrightarrow xyz\left(\dfrac{1}{xy}+\dfrac{1}{xz}+\dfrac{1}{yz}\right)=0\Leftrightarrow x+y+z=0\)
Lại có: \(x^3+y^3+z^3=x^3+y^3+3x^2y+3y^2x-3xy\left(x+y\right)+z^3\)
\(=\left(x+y\right)^3+z^3-3xy\left(-z\right)=\left(x+y+z\right)\left(\left(x+y\right)^2-\left(x+y\right)z+z^2\right)+3xyz=3xyz\)
Vậy nếu \(x+y+z=0\) thì \(x^3+y^3+z^3=3xyz\)
\(P=\dfrac{x^2}{yz}+\dfrac{y^2}{xz}+\dfrac{z^2}{xy}=\dfrac{x^3}{xyz}+\dfrac{y^3}{xyz}+\dfrac{z^3}{xyz}=\dfrac{x^3+y^3+z^3}{xyz}=\dfrac{3xyz}{xyz}=3\)
\(\frac{yz}{x^2}+\frac{xz}{y^2}+\frac{xy}{z^2}=xyz\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)\)
dung hằng đẳng thức đẹp :\(x^3+y^3+z^3=3xyz\) với \(x+y+z=0\)
\(\Rightarrow xyz\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)=xyz\frac{3}{xyz}=3\)
Ta có : \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\Rightarrow\hept{\begin{cases}1+\frac{x}{y}+\frac{x}{z}=0\\\frac{y}{x}+1+\frac{y}{z}=0\\\frac{z}{x}+\frac{z}{y}+1=0\end{cases}}\)
\(\Rightarrow\frac{x}{y}+\frac{x}{z}+\frac{y}{x}+\frac{y}{z}+\frac{z}{x}+\frac{z}{y}=-3\)
mà \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\Rightarrow\frac{yz+xz+xy}{xyz}=0\)
\(\Rightarrow yz+xz+xy=0\)
\(\Rightarrow\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\left(yz+xz+xy\right)=0\)
\(\Rightarrow\frac{yz}{x^2}+\frac{xz}{y^2}+\frac{xy}{z^2}+\frac{x}{y}+\frac{x}{z}+\frac{y}{x}+\frac{y}{z}+\frac{z}{x}+\frac{z}{y}=0\)
\(\Rightarrow\frac{yz}{x^2}+\frac{xz}{y^2}+\frac{xy}{z^2}=3\)
\(\Rightarrow\frac{xy}{z^2}+\frac{yz}{x^2}+\frac{xz}{y^2}=3\)
Học tốt
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
<=> \(\frac{1}{x}+\frac{1}{y}=-\frac{1}{z}\)
<=> \(\left(\frac{1}{x}+\frac{1}{y}\right)^3=\left(-\frac{1}{z}\right)^3\)
<=> \(\frac{1}{x^3}+\frac{1}{y^3}+\frac{3}{x^2y}+\frac{3}{xy^2}=-\frac{1}{z^3}\)
<=> \(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=-\frac{3}{xy}\left(\frac{1}{x}+\frac{1}{y}\right)\)
<=> \(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=-\frac{3}{xy}.\left(-\frac{1}{z}\right)\)
<=> \(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}\)
Khi đó: P = \(\frac{xy}{z^2}+\frac{yz}{x^2}+\frac{xz}{y^2}=\frac{xyz}{z^3}+\frac{xyz}{x^3}+\frac{xyz}{y^3}=xyz.\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)=xyz\cdot\frac{3}{xyz}=3\)
