Đáp án:
`1)x^2+y^2-2x+4y+5=0`
`<=>x^2-2x+1+y^2+4y+4=0`
`<=>(x-1)^2+(y+2)^2=0`
`<=>(x-1)^2=-(y+2)^2`
Vì `(x-1)^2>=0,-(y+2)^2<=0`
`=>(x-1)^2=-(y+2)^2`
`<=>` \(\begin{cases}x-1=0\\y+2=0\\\end{cases}\)
`<=>` \(\begin{cases}x=1\\y=-2\\\end{cases}\)
Vậy `(x,y)=(1,-2)`.
`2)P=x^2+y^2-2x+6y+12`
`=x^2-2x+1+y^2+6y+9+2`
`=(x-1)^2+(y+3)^2+2`
Vì \(\begin{cases}(x-1)^2 \ge 0\\(y+3)^2 \ge 0\\\end{cases}\)
`=>(x-1)^2+(y+3)^2+2>=2`
Hay `P>=2`.
Dấu "=" xảy ra khi \(\begin{cases}x-1=0\\y+3=0\\\end{cases}\)`<=>` \(\begin{cases}x=1\\y=-3\\\end{cases}\)
`3)x+2\sqrt{2x^2}+2x^3=0`
`<=>x+2\sqrt2|x|+2x^3=0`
Nếu `x>=0=>|x|=x`
`pt<=>x+2sqrt2 x+2x^3=0`
`<=>x(1+2sqrt2+x^2)=0`
Vì `x^2+2sqrt2+1>=2sqrt2+1>0AAx`
`<=>x=0`
Nếu `x<=0=>|x|=-x`
`pt<=>x-2sqrt2 x+2x^3=0`
`<=>x(1-2sqrt2+x^2)=0`
`<=>` \(\left[ \begin{array}{l}x=0\\x^2=2\sqrt2-1\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=0\\x=\sqrt{2\sqrt2-1}(l)\\x=-\sqrt{2\sqrt2-1}(tm)\end{array} \right.\)
Vậy `S={0,-\sqrt{2sqrt2-1}}`.
