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2. Tu he co $x,y>0 (1)$$hpt<=>\begin{cases}3yx^2=x^2+2 \\ 3xy^2=y^2+2 \end{cases}$ $=>3(x-y)(3xy+x+y)=0$ $<=>x=y do (1)$Tu do de dang co $x=y=1$4. Tru 2 ve pt co $\sqrt{x^2+9}-\sqrt{y^2+9}+y-x=0$$<=>\frac{x^2-y^2}{\sqrt{x^2+9}+\sqrt{y^2+9}}+y-x=0$$<=>(x-y)(\frac{x+y}{\sqrt{x^2+9}+\sqrt{y^2+9}-1}=0$Vi $x<\sqrt{x^2+9} va y<\sqrt{y^2+9}$Nen $x+y<\sqrt{x^2+9}+\sqrt{y^2+9}=>\frac{x+y}{\sqrt{x^2+9}+\sqrt{y^2+9}}<1$Tu do suy ra $x=y$De dang tim dk $x=y=4$

2. Tu he co $x,y>0 (1)$$hpt<=>\begin{cases}3yx^2=x^2+2 \\ 3xy^2=y^2+2 \end{cases}$ $=>3(x-y)(3xy+x+y)=0$ $<=>x=y do (1)$Tu do de dang co $x=y=1$4. Tru 2 ve pt co $\sqrt{x^2+9}-\sqrt{y^2+9}+y-x=0$$<=>\frac{x^2-y^2}{\sqrt{x^2+9}+\sqrt{y^2+9}}+y-x=0$$<=>(x-y)(\frac{x+y}{\sqrt{x^2+9}+\sqrt{y^2+9}-1})=0$Vi$x<\sqrt{x^2+9} va y<\sqrt{y^2+9}$Nen$x+y<\sqrt{x^2+9}+\sqrt{y^2+9}=>\frac{x+y}{\sqrt{x^2+9}+\sqrt{y^2+9}}<1$Tu do suy ra$x=y$De dang tim dk$x=y=4$