Đk $x\ge 1$$pt(1)\Leftrightarrow x\sqrt{(x+1)^2+(x-y)}+\sqrt{(y+1)^2+(x-y)}=(x+1)^2-(x-y) \quad (*)$Ta có $x\ge y $ vì Nếu $x $(*)\Leftrightarrow \Bigg[(x+1)\sqrt{(x+1)^2+(x-y)}-(x+1)^2 \Bigg]+(x-y)+\Bigg[\sqrt{(y+1)^2+(x-y)}-\sqrt{(x+1)^2+(x-y)^2} \Bigg]=0$$\Leftrightarrow (x-y)\Bigg[\tfrac{x+1}{\sqrt{(x+1)^2+(x-y)}+(x+1)}+1-\tfrac{x+y+2}{\sqrt{(y+1)^2+(x-y)}+\sqrt{(x+1)^2+(x-y)}} \Bigg]=0$$\Leftrightarrow (x-y).A=0\Leftrightarrow x=y$ ($A>0$ do $x\ge y$)Thế $x=y$ vào $pt(2):\sqrt{2(x^2+1)}+\sqrt{x^3+x-2}=1+\sqrt{-(x-1)(2x^2-x+4)+1}$Với đk $x\ge 1$,Dễ thấy $VT \ge 2,VP \le 2$Do đó $VT=VP=1$Có dấu bằng xảy ra khi $x=1$Nghiệm : $(x,y)=(1,1)$
Đk $x\ge 1$$pt(1)\Leftrightarrow x\sqrt{(x+1)^2+(x-y)}+\sqrt{(y+1)^2+(x-y)}=(x+1)^2-(x-y) \quad (*)$Ta có $x\ge y $ vì Nếu $x <y,VT(*) <x(x+1)+(y+1)=VP(*)$$(*)\Leftrightarrow \Bigg[(x+1)\sqrt{(x+1)^2+(x-y)}-(x+1)^2 \Bigg]+(x-y)+\Bigg[\sqrt{(y+1)^2+(x-y)}-\sqrt{(x+1)^2+(x-y)^2} \Bigg]=0$$\Leftrightarrow (x-y)\Bigg[\tfrac{x+1}{\sqrt{(x+1)^2+(x-y)}+(x+1)}+1-\tfrac{x+y+2}{\sqrt{(y+1)^2+(x-y)}+\sqrt{(x+1)^2+(x-y)}} \Bigg]=0$$\Leftrightarrow (x-y).A=0\Leftrightarrow x=y$ ($A>0$ do $x\ge y$)Thế $x=y$ vào $pt(2):\sqrt{2(x^2+1)}+\sqrt{x^3+x-2}=\sqrt{-(x-1)(2x^2-x+4)+1}$Với đk $x\ge 1$,Dễ thấy $VT \ge 1,VP \le 1$Do đó $VT=VP=1$Có dấu bằng xảy ra khi $x=1$Nghiệm : $(x,y)=(1,1)$
Đk $x\ge 1$$pt(1)\Leftrightarrow x\sqrt{(x+1)^2+(x-y)}+\sqrt{(y+1)^2+(x-y)}=(x+1)^2-(x-y) \quad (*)$Ta có $x\ge y $ vì Nếu $x $(*)\Leftrightarrow \Bigg[(x+1)\sqrt{(x+1)^2+(x-y)}-(x+1)^2 \Bigg]+(x-y)+\Bigg[\sqrt{(y+1)^2+(x-y)}-\sqrt{(x+1)^2+(x-y)^2} \Bigg]=0$$\Leftrightarrow (x-y)\Bigg[\tfrac{x+1}{\sqrt{(x+1)^2+(x-y)}+(x+1)}+1-\tfrac{x+y+2}{\sqrt{(y+1)^2+(x-y)}+\sqrt{(x+1)^2+(x-y)}} \Bigg]=0$$\Leftrightarrow (x-y).A=0\Leftrightarrow x=y$ ($A>0$ do $x\ge y$)Thế $x=y$ vào $pt(2):\sqrt{2(x^2+1)}+\sqrt{x^3+x-2}=
1+\sqrt{-(x-1)(2x^2-x+4)+1}$Với đk $x\ge 1$,Dễ thấy $VT \ge
2,VP \le
2$Do đó $VT=VP=1$Có dấu bằng xảy ra khi $x=1$Nghiệm : $(x,y)=(1,1)$