jup e bai nua nha mn: $2x^2+3y^2 \leq 5$. tim max B,min B.biet $B=2x+3y$
mjk thi thay cung bt luc thi thic luc thi get kinh khung.hihi –  damthuy17 05-11-12 10:54 PM
dạng này hay, mình thích –  banhquykeomut 05-11-12 09:07 PM
mình ko thích dạng này –  babylionneu 05-11-12 08:42 PM
a sua gium e hoai nhi –  damthuy17 03-11-12 10:48 PM
Áp dụng bất đẳng thức Bunhia ta có:
$B^2=(2x+3y)^2$
      $=(\sqrt2.\sqrt{2}x+\sqrt3.\sqrt3y)^2\le(2+3)(2x^2+3y^2)\le25$
Suy ra: $-5\le B\le5$
Min$B=-5\Leftrightarrow x=y=-1$
Max$B=5\Leftrightarrow x=y=1$
cong nhan hi –  damthuy17 05-11-12 10:55 PM
cảm ơn lời giải hay –  cuungonghinh 05-11-12 09:23 PM
hay nhỉ, ngắn gọn –  babylionneu 05-11-12 08:42 PM
lời giải hay lắm ạ. ^^ –  duachua.no1 05-11-12 08:24 PM
tks ban nha.jup mjk nhju qua.ban la girl hay boy zay? –  damthuy17 03-11-12 10:41 PM

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