Đáp án:
`2x^2+6\sqrt{2}x+10`
`=2(x^2+3\sqrt{2}x)+10`
`=2[x^2+2*x*3/\sqrt{2}+(3/\sqrt2)^2-(3/\sqrt2)^2]+10`
`=2(x^2+2*x*3/\sqrt{2}+9/2-9/2)+10`
`=2(x^2+2*x*3/\sqrt{2}+9/2)-9+10`
`=2(x+3/\sqrt2)^2+1`
Vì `2(x+3/\sqrt2)^2>=0`
`=>2(x+3/\sqrt2)^2+1>=1`
Hay `2x^2+6\sqrt{2}x+10>=1`(đpcm).
