Đáp án:
Giải thích các bước giải:
\(x^2+xy+y^2+1\)
\(=x^2+2.x.\dfrac{y}{2}+\left(\dfrac{y}{2}\right)^2+\dfrac{3y^2}{4}+1\)
\(=\left(x^2+2.x.\dfrac{y}{2}+\left(\dfrac{y}{2}\right)^2\right)+\dfrac{3y^2}{4}+1\)
\(=\left(x+\dfrac{y}{2}\right)^2+\dfrac{3y^2}{4}+1\ge1>0;\forall x,y\)
Vậy `x^2 + xy + y^2 + 1 > 0 \forall x,y`