Toán 8, đại số: Tìm min, max, nâng cao

N

noinhobinhyen

1.a,

$A=\dfrac{2x^2-16x+41}{x^2-8x+22}=\dfrac{2(x^2-8x+22)-3}{x^2-8x+22}$

$\Leftrightarrow A=2-\dfrac{3}{(x-4)^2+6} \geq 2-\dfrac{3}{6}=\dfrac{3}{2}$

Vậy $min_A=\dfrac{3}{2}\Leftrightarrow x=4$
 
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