ĐK: $x\geq -\frac{1}{8000}$Đặt $2y-1=\sqrt{1+8000x}\Rightarrow \begin{cases}4y^2-4y+1=1+8000x \\ x^2-x-1000(2y-1)=1000 \end{cases}$$\Leftrightarrow \begin{cases}y^2-y-2000x=0 \\ x^2-x-2000y=0 \end{cases}$$\Leftrightarrow (y-x)(y+x)+1999(y-x)=0$$\Leftrightarrow (y-x)(y+x+1999)=0$$\Leftrightarrow \left[ {\begin{matrix} x=y\\ y=-1990-x \end{matrix}} \right.$Với $x=y\Leftrightarrow \begin{cases}x\geq \frac{1}{2} \\ 4x^2-8004x=0 \end{cases}\Leftrightarrow x=2001$Với $y=-1999-x\Leftrightarrow \sqrt{1+8000x}=-3999-2x$ (VN)
ĐK: $x\geq -\frac{1}{8000}$Đặt $2y-1=\sqrt{1+8000x}\Rightarrow \begin{cases}4y^2-4y+1=1+800x \\ x^2-x-1000(2y-1)=1000 \end{cases}$$\Leftrightarrow \begin{cases}y^2-y-2000x=0 \\ x^2-x-2000y=0 \end{cases}$$\Leftrightarrow (y-x)(y+x)+1999(y-x)=0$$\Leftrightarrow (y-x)(y+x+1999)=0$$\Leftrightarrow \left[ {\begin{matrix} x=y\\ y=-1990-x \end{matrix}} \right.$Với $x=y\Leftrightarrow \begin{cases}x\geq \frac{1}{2} \\ 4x^2-8004x=0 \end{cases}\Leftrightarrow x=2001$Với $y=-1999-x\Leftrightarrow \sqrt{1+8000x}=-3999-2x$ (VN)
ĐK: $x\geq -\frac{1}{8000}$Đặt $2y-1=\sqrt{1+8000x}\Rightarrow \begin{cases}4y^2-4y+1=1+800
0x \\ x^2-x-1000(2y-1)=1000 \end{cases}$$\Leftrightarrow \begin{cases}y^2-y-2000x=0 \\ x^2-x-2000y=0 \end{cases}$$\Leftrightarrow (y-x)(y+x)+1999(y-x)=0$$\Leftrightarrow (y-x)(y+x+1999)=0$$\Leftrightarrow \left[ {\begin{matrix} x=y\\ y=-1990-x \end{matrix}} \right.$Với $x=y\Leftrightarrow \begin{cases}x\geq \frac{1}{2} \\ 4x^2-8004x=0 \end{cases}\Leftrightarrow x=2001$Với $y=-1999-x\Leftrightarrow \sqrt{1+8000x}=-3999-2x$ (VN)