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\(A=\left(xy+yz+xz\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-xyz\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\right)\\ =y+x+\dfrac{xy}{z}+y+z+\dfrac{yz}{x}+x+z+\dfrac{xz}{y}-\left(\dfrac{yz}{x}+\dfrac{xz}{y}+\dfrac{xy}{z}\right)\\ =2\left(x+y+z\right)=2.2018=4036\)
Từ \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)\(\Rightarrow\left\{{}\begin{matrix}1+\dfrac{x}{y}+\dfrac{x}{z}=0\left(1\right)\\1+\dfrac{y}{x}+\dfrac{y}{z}=0\left(2\right)\\1+\dfrac{z}{x}+\dfrac{z}{y}=0\left(3\right)\end{matrix}\right.\)
Và \(\dfrac{xy+yz+xz}{xyz}=0\Rightarrow xy+yz+xz=0\)
\(\Rightarrow\left(xy+yz+xz\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\right)=0\)
\(\Rightarrow\dfrac{xy}{z^2}+\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{x}{y}+\dfrac{y}{x}+\dfrac{x}{z}+\dfrac{z}{x}+\dfrac{z}{y}+\dfrac{y}{z}=0\)
\(\Rightarrow A+\dfrac{x}{y}+\dfrac{y}{x}+\dfrac{x}{z}+\dfrac{z}{x}+\dfrac{z}{y}+\dfrac{y}{z}=0\)
Cộng theo vế của \(\left(1\right);\left(2\right);\left(3\right)\)suy ra:
\(\dfrac{x}{y}+\dfrac{y}{x}+\dfrac{y}{z}+\dfrac{z}{y}+\dfrac{z}{x}+\dfrac{x}{z}=-3\)
\(\Rightarrow A-3=0\Rightarrow A=3\)
\(\left\{{}\begin{matrix}xy+yz+xz=0\\x,y,z\ne0\end{matrix}\right.\) \(\Rightarrow\dfrac{1}{z}+\dfrac{1}{y}+\dfrac{1}{x}=0\)\(\Rightarrow\dfrac{1}{z^3}+\dfrac{1}{y^3}+\dfrac{1}{x^3}=\dfrac{3}{zyz}\)
\(A=\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}=xyz\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)=\dfrac{3xyz}{xyz}=3\)
Ta có: \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)\(\Rightarrow xy+yz+xz=0\)
\(\Rightarrow\left\{{}\begin{matrix}xy=-yz-xz\\yz=-xy-xz\\xz=-xy-xz\end{matrix}\right.\)
\(\Rightarrow\dfrac{yz}{x^2+2yz}=\dfrac{yz}{x^2+yz-xy-xz}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\)
Tương tự:
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{xz}{y^2+2xz}=\dfrac{xz}{\left(x-y\right)\left(x-z\right)}\\\dfrac{xy}{z^2+2xy}=\dfrac{xy}{\left(x-y\right)\left(x-z\right)}\\\dfrac{yz}{x^2+2yz}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\end{matrix}\right.\)
\(\Rightarrow A=\dfrac{xz}{\left(x-y\right)\left(x-z\right)}+\dfrac{xy}{\left(x-y\right)\left(x-z\right)}+\dfrac{yz}{\left(x-y\right)\left(x-z\right)}=\dfrac{xz+xy+yz}{\left(x-y\right)\left(x-z\right)}=\dfrac{0}{\left(x-y\right)\left(x-z\right)}=0\)
Vậy \(A=0.\)
ta có : \(xy+yz+xz=0\Rightarrow\dfrac{xy+yz+xz}{xyz}=0\)
\(\Leftrightarrow\dfrac{1}{z}+\dfrac{1}{x}+\dfrac{1}{y}=0\Rightarrow\dfrac{1}{z}=-\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
\(\Rightarrow\dfrac{1}{z^3}=-\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^3\)
\(\Rightarrow\dfrac{1}{z^3}=-\left(\dfrac{1}{x^3}+3.\dfrac{1}{x^2}.\dfrac{1}{y}+3.\dfrac{1}{x}.\dfrac{1}{y^2}+\dfrac{1}{y^3}\right)\)
\(\Rightarrow\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=-3.\dfrac{1}{x}.\dfrac{1}{y}.\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
\(\Rightarrow\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=3.\dfrac{1}{xyz}\)
Do đó : \(xyz.\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)=3\)
\(\Leftrightarrow\dfrac{xyz}{x^3}+\dfrac{xyz}{y^3}+\dfrac{xyz}{z^3}=3\)
\(\Leftrightarrow\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}=3\)
Vậy giá trị của biểu thức \(\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}=3\)
Vì \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Rightarrow\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=\dfrac{3}{xyz}\)(*)
Ta có : \(A=\dfrac{yz}{x^2}+\dfrac{zx}{y^2}+\dfrac{xy}{z^2}=\dfrac{xyz}{x^3}+\dfrac{xyz}{y^3}+\dfrac{xyz}{z^3}=xyz\left(\dfrac{1}{x^3}+\dfrac{1}{x^3}+\dfrac{1}{z^3}\right)\)
\(\Rightarrow A=xyz\left(\dfrac{3}{xyz}\right)=3\)
Lời giải:
Từ \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Rightarrow \frac{xy+yz+xz}{xyz}=0\Rightarrow xy+yz+xz=0\)
Suy ra \(yz=-xy-xz\)
\(\Rightarrow x^2+2yz=x^2+yz-xy-xz=x(x-y)-z(x-y)\)
\(\Leftrightarrow x^2+2yz=(x-z)(x-y)\)
\(\Rightarrow \frac{yz}{x^2+2yz}=\frac{yz}{(x-z)(x-y)}\)
Hoàn toàn tương tự với các phân thức còn lại và cộng theo vế:
\(A=\frac{yz}{(x-y)(x-z)}+\frac{xz}{(y-x)(y-z)}+\frac{xy}{(z-x)(z-y)}\)
\(A=\frac{-yz(y-z)}{(x-y)(y-z)(z-x)}+\frac{-xz(z-x)}{(x-y)(y-z)(z-x)}+\frac{-xy(x-y)}{x-y)(y-z)(z-x)}\)
\(A=\frac{xy^2+yz^2+zx^2-(x^2y+y^2z+z^2x)}{(x-y)(y-z)(z-x)}\)
\(A=\frac{xy^2+yz^2+zx^2-(x^2y+y^2z+z^2x)}{xy^2+yz^2+zx^2-(x^2y+y^2z+z^2x)}=1\)
1)
\(A=\left(x-y+1\right)^2+\left(y-2\right)^2+5\ge5\)
GTNN A=5 khi y=2 và x=1
2)
\(x+y+z=0\Rightarrow x^3+y^3+z^3=3xyz\)
\(A=\dfrac{x^3+y^3+z^3}{xyz}=\dfrac{3xyz}{xyz}=3\)