Đáp án:
`2x+y^2-2ysqrtx-3(2sqrtx-3)=0`
`<=>y^2-2ysqrtx+2x-6sqrtx+9=0`
`<=>y^2-2ysqrtx+x+x-6sqrtx+9=0`
`<=>(y-sqrtx)^2+(sqrtx-3)^2=0`
Vì \(\begin{cases}(y-\sqrt{x})^2 \ge 0\\(\sqrt{x}-3)^2 \ge 0\\\end{cases}\)
`=>(y-sqrtx)^2+(sqrtx-3)^2>=0`
`\text{Mà đề bài cho:}(y-sqrtx)^2+(sqrtx-3)^2=0`
`<=>` \(\begin{cases}y=\sqrt{x}\\\sqrt{x}=3\\\end{cases}\)\
`<=>` \(\begin{cases}y=3\\x=9\\\end{cases}\)
`2)` ta có:(`x+sqrt{x^2+2020})(sqrt{x^2+2020}-x)`
`=x^2+2020-x^2=2020`
Mà `(x+sqrt{x^2+2020})(y+sqrt{y^2+2020})=2020`
`=>y+sqrt{y^2+2020}=sqrt{x^2+2020}-x`
`<=>x+y=sqrt{x^2+2020}-sqrt{y^2+2020}(1)`
Hoàn toàn tương tự ta có:
`x+y=sqrt{y^2+2020}-sqrt{x^2+2020}(2)`
Cộng từng vế (1)(2) ta có:
`2(x+y)=0`
`<=>x+y=0`