\(\dfrac{1}{x}-\dfrac{1}{y}=\dfrac{1}{x}.\dfrac{1}{y}\)

\(\Rightarrow\dfrac{y-x}{xy}=\dfrac{1}{xy}\)

\(\Rightarrow y-x=1\)

\(\Rightarrow y=x+1\)

Vậy S={x,y∈N*\(|y=x+1\)}

(x+y+z)^2=x^2+y^2+z^2

=>2(xy+yz+xz)=0

=>xy+xz+yz=0

=>xy/xyz+xz/xyz+yz/xyz=0

=>1/x+1/y+1/z=0

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{2022}\)

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

\(\Rightarrow\left(yz+zx+xy\right)\left(x+y+z\right)=xyz\)

\(\Rightarrow xy\left(x+y\right)+yz\left(y+z\right)+zx\left(z+x\right)+3xyz-xyz=0\)

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

\(\Rightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)

\(\Rightarrow x=-y\) hoặc \(y=-z\) hoặc \(z=-x\).

-Đến đây thôi bạn, câu hỏi sai rồi ạ.

(x+y+z)^2=x^2+y^2+z^2

=>x^2+y^2+z^2+2(xy+yz+xz)=x^2+y^2+z^2

=>2(xy+yz+xz)=0

=>xy+yz+xz=0

1/x+1/y+1/z

=(xz+yz+xy)/xyz

=0/xyz=0

Có VT = \(\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}}=\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2-\dfrac{2}{xy}-\dfrac{2}{yz}-\dfrac{2}{zx}}\)

\(=\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2-\dfrac{2}{xyz}\left(x+y+z\right)}\) 

\(=\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}=\left|\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right|=VP\) (Vì x + y + z = 0) 

Ta có:\(x:y:z=1:2:3\Rightarrow x=\frac{y}{2}=\frac{z}{3}\).Đặt \(x=\frac{y}{2}=\frac{z}{3}=k\)

\(\Rightarrow\hept{\begin{cases}x=k\\y=2k\\z=3k\end{cases}}\)\(\Rightarrow\left(x+y+z\right)\left(\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\right)=\left(k+2k+3k\right)\left(\frac{1}{k}+\frac{4}{2k}+\frac{9}{3k}\right)\)

\(=6k.\left(\frac{1}{k}+\frac{2}{k}+\frac{3}{k}\right)=6k.\frac{6}{k}=36\)

\(\Rightarrowđpcm\)