Question

Tìm GTNN của biểu thức :

A= 2x² + y² + 2x - 2y +1

Answer 1

`A=2x^2+y^2+2x-2y+1`

`A=2x^2+2x+1/2+y^2-2y+1-1/2`

`A=2(x+1/2)^2+(y-1)^2-1/2`

Vì `2(x+1/2)^2+(y-1)^2 >= 0 AA x,y`

`<=>2(x+1/2)^2+(y-1)^2-1/2 >= -1/2 AA x,y`

  Hay `A >= -1/2 AA x,y`

Dấu "`=`" xảy ra `<=>{(x+1/2=0),(y-1=0):}<=>{(x=-1/2),(y=1):}`

Answer 2

A = 2x² + y² + 2x - 2y + 1

A = [ ( \(\sqrt{2}\) x )² + 2.\(\sqrt{2}\) x . \(\dfrac{1}{\sqrt{2}}\) + ( \(\dfrac{1}{\sqrt{2}}\))² ] + ( y²  - 2y + 1) - \(\dfrac{1}{2}\)

A =   ( \(\sqrt{2}\) x  + \(\dfrac{1}{\sqrt{2}}\) )² + ( y - 1)²  - \(\dfrac{1}{2}\)

vì    ( \(\sqrt{2}\) x + \(\dfrac{1}{\sqrt{2}}\) )² \(\ge\)  0 ; ( y -1 )² \(\ge\) 0 \(\Rightarrow\) A \(\ge\) - \(\dfrac{1}{2}\)

A (min) = - \(\dfrac{1}{2}\) dấu bằng xảy ra

 \(\Leftrightarrow\) \(\left\{{}\begin{matrix}\sqrt{2}x+\dfrac{1}{\sqrt{2}}=0\\y-1=0\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}\sqrt{2}x=-\dfrac{1}{\sqrt{2}}\\y=1\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=1\end{matrix}\right.\)