tìm min max

N

nhungnguoithaniu

hì hì

[tex]y = 3\sqrt {x - 1} + 4\sqrt {5 - x} [/tex]
[tex]|y| \le \sqrt {{3^2} + {4^2}} \sqrt {x - 1 + 5 - x} = 10(bdt - bunhiacopsky)[/tex]
[tex] \Rightarrow - 10 \le y \le 10[/tex]
[tex]dau - bang \Leftrightarrow 4\sqrt {x - 1} = 3\sqrt {5 - x} \Leftrightarrow ...[/tex]
[tex] \Rightarrow \min ,\max [/tex]
 
V

vy000

$y^2=9(x-1)+16(5-x)+2.3\sqrt{x-1}.4\sqrt{5-x} \\ \ge 9x-9+80-16x \\ \ge 71-7x \\ \ge 71-35 \\ =36$

Vì $y \ge 0 $ \Leftrightarrow $y \ge 6$
 
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