Bài 5. (0,5 điểm). Học sinh chọn 1 trong hai đề a) Giải hệ phương trình $\left \{ {{x^2+y^2= 3-xy} \atop {x^4 + y^4= 0}} \right.$  b) b) Let x, y $\in$R, x≥0; y $\geq$ 0 satisfy x+y=2, find the maximum and minimum of expression: H = $\sqrt{x^2+ xy+y^2}$ + $\sqrt{3xy}$

** Max:

Ta có:

`H=\sqrt(x^2+xy+y^2)+\sqrt(3xy)`

`H^2=(1.\sqrt(x^2+xy+y^2)+1.\sqrt(3xy))^2`

`H^2<=(1^2+1^2)(x^2+xy+y^2+3xy)`

`H^2<=2(x^2+2xy+y^2+2xy)`

`H^2<=2[(x+y)^2+2*(x+y)^2/4]=2(2^2+2*2^2/4)=12`

`H<=\sqrt(12)=2\sqrt(3)`

Dấu "=" xảy ra khi: `x=y=1`

** Min:

Ta có:

`H=\sqrt(x^2+xy+y^2)+\sqrt(3xy)`

`H^2=x^2+xy+y^2+3xy+2\sqrt((x^2+xy+y^2)*3xy)`

`H^2=(x^2+2xy+y^2)+2xy+2\sqrt((x^2+xy+y^2)*3xy)`

`H^2=(x+y)^2+2xy+2\sqrt((x^2+xy+y^2)*3xy)`

`H^2>=2^2+2*0*y+2\sqrt((0^2+0*y+y^2)*3*0*y)`

`H^2>=4`

`H>=2`

Dấu "=" xảy ra khi: `(x,y)\in{(0,2);(2,0)}`