Question

[Tam Nguyen]

Cm rằng : A= x² + y² - 4x +2y +7 > 0 với mọi số thực x, y

Answer 1

[Lê Thiên Anh]

A= x²+y²-4x+2y+7

= (x²-4x+4)+(y²+2y+1)+2

= (x-2)²+(y+1)²+2

Ta thấy: (x-2)²\(\ge0\)

(y+1)²\(\ge0\)

\(\Rightarrow\)(x-2)²+(y+1)²+2\(\ge2\)

\(\Rightarrow\)A\(\ge2\)

Vậy A>0 \(\forall x,y\)

Answer 2

[Lightning Farron]

\(A=x^2+y^2-4x+2y+7\)

\(=x^2+y^2-4x+2y+4+1+2\)

\(=\left(x^2-4x+4\right)+\left(y^2+2y+1\right)+2\)

\(=\left(x-2\right)^2+\left(y+1\right)^2+2\)

Ta thấy: \(\left\{{}\begin{matrix}\left(x-2\right)^2\ge0\forall x\\\left(y+1\right)^2\ge0\forall y\end{matrix}\right.\)

\(\Rightarrow\left(x-2\right)^2+\left(y+1\right)^2\ge0\forall x,y\)

\(\Rightarrow\left(x-2\right)^2+\left(y+1\right)^2+2\ge2>0\forall x,y\)