Question

[Tiến Hoàng Minh]

Tìm GTLN GTNN của A=\(\dfrac{\text{ 2x+1}}{\text{x^2+2 }}\)

Answer 1

[Minh Hiếu]

\(A=\dfrac{2x+1}{x^2+2}\)

\(\Leftrightarrow Ax^{2\:}+2A=2x+1\)

+) \(A=0\Rightarrow x=-\dfrac{1}{2}\)

+) \(A\ne0\)

\(Ax^2+2A=2x+1\)

\(\Leftrightarrow Ax^{2\:}-2x=1-2A\)

\(\Leftrightarrow x^2-2.\dfrac{x}{A}=\dfrac{1-2A}{A}\)

\(\Leftrightarrow x^2-2.x.\dfrac{1}{A}+\dfrac{1}{A^2}=\dfrac{1-2A}{A}+\dfrac{1}{A^2}\)

\(\Leftrightarrow\left(x-\dfrac{1}{A}\right)^2=\dfrac{A-2A^2+1}{A^2}\)

\(\Leftrightarrow\left(x-\dfrac{1}{A}\right)^2=\dfrac{\left(1-A\right)\left(2A+1\right)}{A^2}\)

Vì \(\left\{{}\begin{matrix}\left(x-\dfrac{1}{A}\right)^2\ge0\left(\forall x,A\ne0\right)\\A^2\ge0\end{matrix}\right.\)

⇒ \(\left(1-A\right)\left(2A+1\right)\ge0\)

⇒ \(-\dfrac{1}{2}\le A\le1\)

Còn lại tụ làm nha

Answer 2

[☆Châuuu~~~(๑╹ω╹๑ )☆]

\(A=\dfrac{2x+1}{x^2+2}=\dfrac{x^2+2-x^2-2+2x+1}{x^2+2}\\ =1-\dfrac{-\left(x-1\right)^2}{x^2+2}\\ Do\left(x-1\right)^2\ge0\Rightarrow\dfrac{-\left(x-1\right)^2}{x^2+2}\ge0\\ \Rightarrow\dfrac{-\left(x-1\right)^2}{x^2+2}=0\Leftrightarrow\dfrac{-\left(x-1\right)^2}{x^2+2}+1\le1\) 

\(Dấu"="\Leftrightarrow A=1\\ \Leftrightarrow x-1=0\Rightarrow x=1\\ Vậy.P_{max}=1.khi.x=1\\ A=\dfrac{2x+1}{x^2+2}\rightarrow2A+1=\dfrac{2.\left(2x+1\right)}{x^2+2}+1\\ =\dfrac{4x+2+x^2+2}{x^2+2}=\dfrac{x^2+4x+2}{x^2+2}=\dfrac{\left(x+2\right)^2}{x^2+2}\\ Do\left(x+2\right)^2\ge0\Leftrightarrow\dfrac{\left(x+2\right)^2}{x^2+2}\ge0\) 

\(Dấu"="\Leftrightarrow A=\dfrac{1}{2}khi.x=-2\\ \Rightarrow2A+1\ge0\Rightarrow2A\ge-1\Rightarrow A>-\dfrac{1}{2}\\ Vậy.MinA=-\dfrac{1}{2}.khi.x=-2\)