\(A=\dfrac{x^2+x+1}{\left(x-1\right)^2}\\ ĐKXĐ:x\ne1\\ \Rightarrow A=\dfrac{x^2-2x+3x+1+3-3}{\left(x-1\right)^2}\\ =\dfrac{\left(x^2-2x+1\right)+\left(3x-3\right)+3}{\left(x-1\right)^2}\\ =\dfrac{x^2-2x+1}{\left(x-1\right)^2}+\dfrac{3x-3}{\left(x-1\right)^2}+\dfrac{3}{\left(x-1\right)^2}\\ =\dfrac{\left(x-1\right)^2}{\left(x-1\right)^2}+\dfrac{3\left(x-1\right)}{\left(x-1\right)^2}+\dfrac{3}{\left(x-1\right)^2}\\ =1+\dfrac{3}{x-1}+\dfrac{3}{\left(x-1\right)^2}\)
Đặt \(\dfrac{1}{x-1}=t\)
\(\Rightarrow A=3t^2+3t+1\\ =3t^2+3t+\dfrac{3}{4}+\dfrac{1}{4}\\ =\left(3t^2+3t+\dfrac{3}{4}\right)+\dfrac{1}{4}\\ =3\left(t^2+t+\dfrac{1}{4}\right)+\dfrac{1}{4}\\ =3\left(t+\dfrac{1}{2}\right)^2+\dfrac{1}{4}\)
Do \(3\left(t+\dfrac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow P=3\left(t+\dfrac{1}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\forall x\)
Dấu "=" xảy ra khi:
\(3\left(t+\dfrac{1}{2}\right)^2=0\\ \Leftrightarrow t+\dfrac{1}{2}=0\\ \Leftrightarrow t=-\dfrac{1}{2}\\ \Leftrightarrow\dfrac{1}{x-1}=-\dfrac{1}{2}\\ \Leftrightarrow x-1=-2\\ \Leftrightarrow x=-1\)
Vậy \(A_{Min}=\dfrac{1}{4}\) khi \(x=-1\)
Cách khác :
\(A=\dfrac{x^2+x+1}{\left(x-1\right)^2}\) ( x # 1)
\(A=\dfrac{4\left(x^2+x+1\right)}{4\left(x-1\right)^2}\)
\(A=\dfrac{x^2-2x+1+3x^2+6x+3}{4\left(x-1\right)^2}\)
\(A=\dfrac{\left(x-1\right)^2}{4\left(x-1\right)^2}+\dfrac{3\left(x^2+2x+1\right)}{4\left(x-1\right)^2}\)
\(A=\dfrac{1}{4}+\dfrac{3\left(x+1\right)^2}{4\left(x-1\right)^2}\)
Do : \(\dfrac{3\left(x+1\right)^2}{4\left(x-1\right)^2}\) ≥ 0 ∀x
⇒ \(\dfrac{3\left(x+1\right)^2}{4\left(x-1\right)^2}\) \(+\dfrac{1}{4}\) ≥ \(\dfrac{1}{4}\)
⇒ AMin = \(\dfrac{1}{4}\) ⇔ x = - 1
