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$(1) \Leftrightarrow\sqrt{(x+3y-2)^2+4}+(x+3y-2)=\sqrt{(x-y)^2+4}-(x-y)$Đặt $u=x+3y-2;v=x-y$$\Leftrightarrow \sqrt{u^2+4}+u=\sqrt{v^2+4}-v$$\Leftrightarrow (u+v)(\frac{u+v}{\sqrt{u^2+v^2}}-1)=0$$\Leftrightarrow (u-v)(u+v-\sqrt{u^2+v^2})=0$$\left[ {\begin{matrix}u=v\\ \begin{cases}u=0 \\ v\ge0\end{cases} \\\begin{cases}u\ge0 \\ v=0 \end{cases}\end{matrix}} \right.$Thay dần vào :v

$(1) \Leftrightarrow\sqrt{(x+3y-2)^2+4}+(x+3y-2)=\sqrt{(x-y)^2+4}-(x-y)$Đặt $u=x+3y-2;v=x-y$$\Leftrightarrow \sqrt{u^2+4}+u=\sqrt{v^2+4}-v$$\Leftrightarrow (u+v)(\frac{u+v}{\sqrt{u^2+v^2}}-1)=0$$\Leftrightarrow (u+v)(u+v-\sqrt{u^2+v^2})=0$$\left[ {\begin{matrix}u=-v\\ \begin{cases}u=0 \\ v\ge0\end{cases} \\\begin{cases}u\ge0 \\ v=0 \end{cases}\end{matrix}} \right.$Thay dần vào :v

$(1) \Leftrightarrow\sqrt{(x+3y-2)^2+4}+(x+3y-2)=\sqrt{(x-y)^2+4}-(x-y)$Đặt $u=x+3y-2;v=x-y$$\Leftrightarrow \sqrt{u^2+4}+u=\sqrt{v^2+4}-v$$\Leftrightarrow (u+v)(\frac{u+v}{\sqrt{u^2+v^2}}+-)=0$$\Leftrightarrow (u-v)(u+v-\sqrt{u^2+v^2})=0$$\left[ {\begin{matrix}u=v\\ \begin{cases}u=0 \\ v\ge0\end{cases} \\\begin{cases}u\ge0 \\ v=0 \end{cases}\end{matrix}} \right.$Thay dần vào :v

$(1) \Leftrightarrow\sqrt{(x+3y-2)^2+4}+(x+3y-2)=\sqrt{(x-y)^2+4}-(x-y)$Đặt $u=x+3y-2;v=x-y$$\Leftrightarrow \sqrt{u^2+4}+u=\sqrt{v^2+4}-v$$\Leftrightarrow (u+v)(\frac{u+v}{\sqrt{u^2+v^2}}-1)=0$$\Leftrightarrow (u-v)(u+v-\sqrt{u^2+v^2})=0$$\left[ {\begin{matrix}u=v\\ \begin{cases}u=0 \\ v\ge0\end{cases} \\\begin{cases}u\ge0 \\ v=0 \end{cases}\end{matrix}} \right.$Thay dần vào :v