+) GTLN :
A = \(\dfrac{2x+1}{x^2+2}\)
A = \(\dfrac{x^2+2-x^2-2x-1}{x^2+2}\)
A = 1 - \(\dfrac{\left(x+1\right)^2}{x^2+2}\)
Do : - \(\dfrac{\left(x+1\right)^2}{x^2+2}\) ≤ 0 ∀x
⇒ 1 - \(\dfrac{\left(x+1\right)^2}{x^2+2}\) ≤ 1
⇒ AMAX = 1 ⇔ x = -1
+) GTNN :
A = \(\dfrac{2x+1}{x^2+2}\)
A = \(\dfrac{-2x^2-4+x^2+4x+4+x^2-2x+1}{x^2+2}\)
A = \(\dfrac{-2\left(x^2+2\right)}{x^2+2}+\dfrac{\left(x+2\right)^2+\left(x-1\right)^2}{x^2+2}\)
A = - 2 + \(\dfrac{\left(x+2\right)^2+\left(x-1\right)^2}{x^2+2}\)
Do : \(\dfrac{\left(x+2\right)^2+\left(x-1\right)^2}{x^2+2}\) ≥ 0 ∀x
⇒ - 2 + \(\dfrac{\left(x+2\right)^2+\left(x-1\right)^2}{x^2+2}\) ≥ - 2
⇒AMAX = - 2 ⇔ x = 1 ; x = -2
\(\text{*}\dfrac{2x+1}{x^2+2}=\dfrac{2x+2-1-x^2+x^2}{x^2+2}\\ =\dfrac{-\left(x^2+2x+1\right)+\left(x^2+2\right)}{x^2+2}\\ =\dfrac{-\left(x^2+2x+1\right)}{x^2+2}+\dfrac{x^2+2}{x^2+2}=\dfrac{-\left(x+1\right)^2}{x^2+2}+1\)
Do \(-\left(x+1\right)^2\le0\forall x\)
\(x^2+2>0\forall x\\ \Rightarrow\dfrac{-\left(x+1\right)^2}{x^2+2}\le0\forall x\\ \Rightarrow\dfrac{-\left(x+1\right)^2}{x^2+2}+1\le1\forall x\)
Dấu "=" xảy ra khi:
\(-\left(x+1\right)^2=0\\ \Leftrightarrow x+1=0\\ \Leftrightarrow x=-1\)
\(\text{*}\dfrac{2x+1}{x^2+2}=\dfrac{4x+2}{2\left(x^2+2\right)}=\dfrac{4x+4-2+x^2-x^2}{2\left(x^2+2\right)}\\ =\dfrac{\left(x^2+4x+4\right)-\left(x^2+2\right)}{2\left(x^2+2\right)}\\ =\dfrac{x^2+4x+4}{2\left(x^2+2\right)}-\dfrac{x^2+2}{2\left(x^2+2\right)}=\dfrac{\left(x+2\right)^2}{2\left(x^2+2\right)}-\dfrac{1}{2}\)
Do \(\left(x+2\right)^2\ge0\forall x\)
\(2\left(x^2+2\right)>0\forall x\\ \Rightarrow\dfrac{\left(x+2\right)^2}{2\left(x^2+2\right)}\ge0\forall x\\ \Rightarrow\dfrac{\left(x+2\right)^2}{2\left(x^2+2\right)}-\dfrac{1}{2}\ge-\dfrac{1}{2}\forall x\)
Dấu "=" xảy ra khi:
\(\left(x+2\right)^2=0\\ \Leftrightarrow x+2=0\\ \Leftrightarrow x=-2\)
Vậy \(GTLN\) của biểu thức là \(1\) khi \(x=-1\)
\(GTNN\) của biểu thức là \(-\dfrac{1}{2}\) khi \(x=-2\)
