\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\sqrt{3}\)

\(\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2=3\)

\(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+\dfrac{2}{xy}+\dfrac{2}{yz}+\dfrac{2}{xz}=3\)

\(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+2.\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}\right)=3\)

\(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+2.\left(\dfrac{x+y+z}{xyz}\right)=3\)

\(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+2.1=3\) ( Do x+y+z=xyz )

\(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}=3-2=1\)

Vậy P = 1

Có x=a/m; y=b/m và x<y nên a/m<b/m ⇒a<b

Giả sử z>x là đúng thì\(\dfrac{a+b}{2m}>\dfrac{a}{m}\Leftrightarrow\dfrac{a+b}{2m}-\dfrac{a}{m}>0\\ \Leftrightarrow\dfrac{a+b-2a}{2m}>0\Leftrightarrow\dfrac{b-a}{2m}>0\\ m\text{à}b>a;m>0n\text{ê}nz>xl\text{à}\text{đ}\text{úng (1)}\)Giả sử z<y là đúng thì

\(\dfrac{a+b}{2m}< \dfrac{b}{m}\Leftrightarrow\dfrac{a+b}{2m}-\dfrac{b}{m}< 0\\ \Leftrightarrow\dfrac{a+b-2b}{2m}< 0\Leftrightarrow\dfrac{a-b}{2m}< 0\\ m\text{à}a< b;m>0n\text{ê}nz< yl\text{à}\text{đ}\text{úng (2)}\)

Từ (1)và(2) suy ra đpcm

x/5=3/y

nên xy=15

mà 0<x<y

nên \(\left(x,y\right)\in\left\{\left(1;15\right);\left(3;5\right)\right\}\)

Vì \(\dfrac{-2}{x}=\dfrac{y}{6}\)

\(\Rightarrow xy=-2.6\)

\(\Rightarrow xy=-12\)

Mà x<0<y

\(\Rightarrow\) x \(\in\) {-1;-2;-3;-4;-6;-12}

\(\Rightarrow\) y \(\in\) {12; 6; 4; 3; 2; 1}

Vậy (x,y)=(-1,12);(-2;6);(-3;4);(-4;3);(-6;2);(-12;1) thỏa mãn bài ra

\(\dfrac{-2}{x}=\dfrac{y}{6}\)

\(\Rightarrow xy=6.-2\)

\(\Rightarrow xy=-12\)

\(\Rightarrow xy\inƯ\left(-12\right)\)

\(Ư\left(-12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)

\(\Rightarrow\left\{{}\begin{matrix}x=1\\y=-12\\x=-1\\y=12\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=2\\y=-6\\x=-2\\y=6\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=3\\y=-4\\x=-3\\y=4\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=4\\y=-3\\x=-4\\y=3\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=6\\y=-2\\x=-6\\y=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=12\\y=-1\\x=-12\\y=1\end{matrix}\right.\)