cho x^2 + 3y^2=4 . tìm min, max của P= x+y
cho x^2 +2y^2 -2xy=5 tìm min , max p= x+y-1
1/ $(x+y)^2=(x+\dfrac{1}{\sqrt3}\sqrt3y)^2\le (1+\dfrac13)(x^2+3y^2)=\dfrac{16}{3}$
$\Rightarrow -\dfrac{4\sqrt3}{3}\le P\le \dfrac{4\sqrt3}{3}$
$P_{min}=-\dfrac{4\sqrt3}{3}\Leftrightarrow \left\{\begin{matrix}x=3y<0\\x^2+3y^2=4\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix}x=\dfrac{-3\sqrt3}{3}\\y=\dfrac{-\sqrt3}{3}\end{matrix}\right.$
$P_{max}=\dfrac{4\sqrt3}{3}\Leftrightarrow \left\{\begin{matrix}x=3y>0\\x^2+3y^2=4\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix}x=\dfrac{3\sqrt3}{3}\\y=\dfrac{\sqrt3}{3}\end{matrix}\right.$
2/ Ta có: $x^2 +2y^2 -2xy=5\Leftrightarrow (x-y)^2+y^2=5$
$(x+y)^2=(x-y+2y)^2\le (1+2^2)[(x-y)^2+y^2]=25$
$\Rightarrow -5\le x+y\le 5$
$\Rightarrow -6\le P\le 4$
$P_{min}=-6 \Leftrightarrow \left\{\begin{matrix}2(x-y)=y<0\\(x-y)^2+y^2=5\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix}x=-3\\y=-2\end{matrix}\right.$
$P_{max}=4 \Leftrightarrow \left\{\begin{matrix}2(x-y)=y>0\\(x-y)^2+y^2=5\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix}x=3\\y=2\end{matrix}\right.$
Có gì khúc mắc em hỏi lại nhé
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
