ĐKXĐ: \(x\in\mathbb{R}\)
Đặt \(A=\dfrac{x+3}{x^2+7}\). Khi đó:
Xét: \(A-\dfrac{1}{2}=\dfrac{x+3}{x^2+7}-\dfrac{1}{2}=\dfrac{2\left(x+3\right)}{2\left(x^2+7\right)}-\dfrac{x^2+7}{2\left(x^2+7\right)}\)
\(=\dfrac{2x+6-x^2-7}{2\left(x^2+7\right)}=\dfrac{-x^2+2x-1}{2\left(x^2+7\right)}=\dfrac{-\left(x-1\right)^2}{2\left(x^2+7\right)}\)
Ta thấy: \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\forall x\\2\left(x^2+7\right)>0\forall x\end{matrix}\right.\Rightarrow\dfrac{\left(x-1\right)^2}{2\left(x^2+7\right)}\ge0\)
\(\Rightarrow\dfrac{-\left(x-1\right)^2}{2\left(x^2+7\right)}\le0\)
\(\Leftrightarrow A-\dfrac{1}{2}\le0\Leftrightarrow A\le\dfrac{1}{2}\)
Dấu \("="\) xảy ra khi: \(x-1=0\Leftrightarrow x=1\)
Vậy GTLN của biểu thức đã cho là \(\dfrac{1}{2}\) tại \(x=1\).
\(\text{#}Toru\)
