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Tìm min các biểu thức sau đây (Giúp mình há)
$A=x^2+y^2-xy-3y+6$
$A=x^2+y^2-xy-3y+6=x^2 + \dfrac{1}{4}y^2 + \dfrac{3}{4}y^2- 2.\dfrac{1}{2}xy-3y+ 3 + 3 $
$=(x^2 + \dfrac{1}{4}y^2 - 2.\dfrac{1}{2}xy)-3.(y+ \dfrac{1}{4}y^2+1)+ 3 $
$=(x^2 + \dfrac{1}{4}y^2 - 2.\dfrac{1}{2}xy)+3.(-y+ \dfrac{1}{4}y^2+1)+ 3 $
$=(x- \dfrac{1}{2}y)^2 +3.( \dfrac{1}{2}y-1)^2+ 3 $
Ta có $(x- \dfrac{1}{2}y)^2+3.( \dfrac{1}{2}y-1)^2 \ge 0 \to (x- \dfrac{1}{2}y)^2 +3.( \dfrac{1}{2}y-1)^2+ 3 \ge 3$ \forall x;y
Dấu "=" xảy ra $\leftrightarrow \left\{\begin{matrix} (x- \dfrac{1}{2}y)^2=0 \\ 3.( \dfrac{1}{2}y-1)^2=0 \end{matrix}\right. \leftrightarrow \left\{\begin{matrix} x- \dfrac{1}{2}y=0 \\ \dfrac{1}{2}y-1=0 \end{matrix}\right. \leftrightarrow \left\{\begin{matrix} x- \dfrac{1}{2}.2=0 \\ y=2 \end{matrix}\right. \leftrightarrow \left\{\begin{matrix} x=1 \\ y=2 \end{matrix}\right.$
Vậy $Min A = 3 \leftrightarrow \left\{\begin{matrix} x=1 \\ y=2 \end{matrix}\right.$
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