Tìm Min

S

soicon_boy_9x

Đặt $y=1-x ( 0<y<1)$. $x=(1-y)$

$\leftrightarrow A=\dfrac{4}{y}+\dfrac{1}{y(1-y)^2}$

$\leftrightarrow A=\dfrac{1}{y}(4+\dfrac{1}{(1-y)^2}\geq 4\dfrac{1}{y(1-y)}$

Tìm max $y(1-y)$

Ta có:

$y(1-y)=y-y^2 = y-y^2-0,25+0,25=-(y-\dfrac{1}{2})^2+0,25 \leq 0$

Dấu = xảy ra khi $y=0,5 \leftrightarrow x=0,5$

Vậy Min A là : $4\dfrac{1}{y(1-y)}=16$


 
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