[Toán 8]tìm giá trị lớn nhất

C

congchuaanhsang

A=$\dfrac{-16x^2-1}{x^2+15}$=$\dfrac{-16(x^2+15)+239}{x^2+15}$

=$\dfrac{239}{x^2+15}-16$

$x^2+15$\geq15 với mọi x\Rightarrow$\dfrac{239}{x^2+15}$\leq$\dfrac{239}{15}$

\RightarrowA\leq$\dfrac{239}{15}-16$=$\dfrac{-1}{15}$

$A_{max}=\dfrac{-1}{15}$\Leftrightarrow$x=0$
 
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