tính giá trị nhỏ nhất của biểu thức:A=x2+10y2+4xy-4x-2y+20
Ta có : A=x2+10y2+4xy-4x-2y+20
→ A = \(\left(x^2+4xy+4y^2\right)-\left(4x+8y\right)+4+\left(6y^2+6y+\dfrac{6}{4}\right)-\dfrac{6}{4}+16\)
→ A = \(\left[\left(x+2y\right)^2-4\left(x+2y\right)+2^2\right]+6\left(y^2+\dfrac{2.y}{2}+\dfrac{1}{4}\right)-\dfrac{6}{4}+16\)
→ A = \(\left(x+2y-2\right)^2+6\left(y+\dfrac{1}{2}\right)^2+\dfrac{29}{2}\) ≥ \(\dfrac{29}{2}\)
Vì \(\left(x+2y-2\right)^2\ge0\) ∀ x,y
\(6\left(y+\dfrac{1}{2}\right)^2\ge0\) ∀ y
⇒ GTNN của A là \(\dfrac{29}{2}\)
Dấu = xảy ra ⇔ \(\left\{{}\begin{matrix}\left(x+2y-2\right)^2=0\\\left(y+\dfrac{1}{2}\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x+2y-2=0\\y+\dfrac{1}{2}=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{1}{2}\\x=3\end{matrix}\right.\)
Vậy A đạt GTNN là \(\dfrac{29}{2}\) \(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{1}{2}\\x=3\end{matrix}\right.\)