hoặc \(x+y+z=-\sqrt{31\frac{6}{17}}\)

=> xy+yz+zx =105/17

Ta có\(\left(x+y+z\right)^2=x^2+y^2+z^2+2\left(xy+yz+zx\right)\)

                                     = 19                               + 2. 105/17

                                     =533/17

=> \(x+y+z=\sqrt{\frac{533}{17}}\)

Lời giải:

\(yz-xz-xy=0\Rightarrow yz-xz=xy\)

\(B=\frac{yz}{x^2}-\frac{zx}{y^2}-\frac{xy}{z^2}\)\(=\frac{(yz)^3-(xz)^3-(xy)^3}{x^2y^2z^2}\)

Xét: \((yz)^3-(xz)^3-(xy)^3=(yz-xz)^3+3yz.xz(yz-xz)-(xy)^3\)

\(=(xy)^3+3yz.xz.xy-(xy)^3=3x^2y^2z^2\)

\(\Rightarrow B=\frac{(yz)^3-(xz)^3-(xy)^3}{x^2y^2z^2}=\frac{3x^2y^2z^2}{x^2y^2z^2}=3\)

\(x+y+z=0\Rightarrow\left(x+y+z\right)^2=0\Rightarrow x^2+y^2+z^2+2xy+2yz+2zx=0\)

\(\Rightarrow x^2+y^2+z^2+2\left(xy+yz+zx\right)=0\)

Mà \(xy+yz+zx=0\)(theo đề) nên \(2\left(xy+yz+zx\right)=0\)

\(\Rightarrow x^2+y^2+z^2=0\)

Vì \(\hept{\begin{cases}x^2\ge0\\y^2\ge0\\z^2\ge0\end{cases}}\) (với mọi x;y;z) nên \(x^2+y^2+z^2\ge0\) (với mọi x;y;z)

Để \(x^2+y^2+z^2=0\) \(\Leftrightarrow\) \(\hept{\begin{cases}x^2=0\\y^2=0\\z^2=0\end{cases}\Leftrightarrow}x=y=z=0\)

Vậy \(A=\left(0-1\right)^{2016}+0^{2017}+\left(0+1\right)^{2018}=\left(-1\right)^{2016}+0+1^{2018}=2\)

cũng bằng 3

Ta coˊ :xy+x+1x​+yz+y+1y​+xz+z+1z​

=���+�+1+�����+��+�+����2��+���+��=xy+x+1x​+xyz+xy+xxy​+x2yz+xyz+xyxyz​

=���+�+1+����+�+1+1��+�+1(Vıˋ ���=1)=xy+x+1x​+xy+x+1xy​+xy+x+11​(Vıˋ xyz=1)

=�+��+1��+�+1=xy+x+1x+xy+1​

$=1$

\(\dfrac{xy}{x+y}=\dfrac{yz}{y+z}=\dfrac{zx}{z+x}\\ \Rightarrow\dfrac{x+y}{xy}=\dfrac{y+z}{yz}=\dfrac{z+x}{zx}\\ \Rightarrow\dfrac{1}{y}+\dfrac{1}{x}=\dfrac{1}{z}+\dfrac{1}{y}=\dfrac{1}{x}+\dfrac{1}{z}\\ \Rightarrow\dfrac{1}{x}=\dfrac{1}{y}=\dfrac{1}{z}\\ \Rightarrow x=y=z\)

\(\Rightarrow P=\dfrac{xy+yz+zx}{x^2+y^2+z^2}=\dfrac{x^2+x^2+x^2}{x^2+x^2+x^2}=1\)