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\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)
\(\Rightarrow xy+yz+xz=0\)
\(\Rightarrow\left\{{}\begin{matrix}xy=-yz--xz\\yz=-xy-xz\\xz=-xy-xz\end{matrix}\right.\)
\(\dfrac{yz}{x^2+2yz}=\dfrac{yz}{x^2+yz-xy-xz}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\)
CMTT:
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{xz}{y^2+2xz}=\dfrac{xz}{\left(x-y\right)\left(x-z\right)}\\\dfrac{xy}{z^2+2xy}=\dfrac{xy}{\left(x-y\right)\left(x-z\right)}\\\dfrac{yz}{x^2+2yz}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\end{matrix}\right.\)
A=\(\dfrac{xz}{\left(x-y\right)\left(x-z\right)}+\dfrac{xy}{\left(x-y\right)\left(x-z\right)}+\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\)
\(A=\dfrac{xz+xy+yz}{\left(x-y\right)\left(x-z\right)}\left(1\right)\)
mà \(xy+yz+xz=0\)
Từ \(\Rightarrow\dfrac{xz+xy+yz}{\left(x-y\right)\left(x-z\right)}=0\)
Vậy A=0
\(xyz\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=0\\ \Rightarrow yz+xz+xy=0\)
\(A=\frac{xy}{z^2}+\frac{xz}{y^2}+\frac{yz}{x^2}\\ \Leftrightarrow A=\frac{x^3y^3+x^3z^3+y^3z^3}{x^2y^2z^2}\)
Ta có :\(yz+xz+xy=0\)
\(\Rightarrow y^3x^3+x^3z^3+x^3y^3=-3xyz\left(y^2z+yz^2+x^2z+xz^2+x^2y+xy^2+2xyz\right)\)
\(=-3xyz\left(yz+xz\right)\left(xz+xy\right)\left(yz+xy\right)\)
\(=-3xyz\left(-xy\right)\left(-yz\right)\left(-xz\right)\\ =3x^2y^2z^2\)
\(\Rightarrow A=\frac{3x^2y^2z^2}{x^2y^2z^2}=3\)
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Rightarrow xy+yz+xz=0\) (nhân 2 vế với\(xyz\ne0\))
=> x2 + 2yz = x2 + 2yz - xy - yz - xz = x2 - xz - xy + yz = x(x - z) - y(x - z) = (x - y)(x - z).
Tương tự,y2 + 2xz = (y - x)(y - z) ; z2 + 2xy = (z - x)(z - y)
\(\Rightarrow\frac{yz}{x^2+2yz}+\frac{xz}{y^2+2xz}+\frac{xy}{z^2+2xy}=\frac{yz}{\left(x-y\right)\left(x-z\right)}+\frac{xz}{\left(y-x\right)\left(y-z\right)}+\frac{xy}{\left(z-x\right)\left(z-y\right)}=1\)
Dễ dàng chứng minh được : nếu \(a+b+c=0\) thì \(a^3+b^3+c^3=3abc\)
Ta có \(\frac{xy}{z^2}+\frac{yz}{x^2}+\frac{zx}{y^2}=xyz\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)=xyz.\frac{3}{xyz}=3\)( Vì \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\))
Ta có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Rightarrow\frac{xy+yz+zx}{xyz}=0\Rightarrow xy+yz+zx=0\)
\(P=\frac{xy}{z^2}+\frac{yz}{x^2}+\frac{xz}{y^2}=\frac{x^3y^3+y^3z^3+x^3z^3}{x^2y^2z^2}\)
Áp dụng nếu a+b+c=0 thì a3+b3+c3=3abc
Với a=xy, b=yz, c=zx
Ta có: \(P=\frac{a^3+b^3+c^3}{abc}=\frac{3abc}{abc}=3\)
Vậy P=3