Lời giải:
Ta có: \(A=\frac{3x^2-2x+3}{x^2+1}\Rightarrow 3x^2-2x+3=(x^2+1)A\)
\(\Leftrightarrow x^2(3-A)-2x+(3-A)=0\)
Vì biểu thức $A$ tồn tại nên pt trên có nghiệm. Tức là:
\(\Delta'=(-1)^2-(3-A)(3-A)\geq 0\)
\(\Leftrightarrow 1-(3-A)^2\geq 0\)
\(\Leftrightarrow (A-2)(4-A)\geq 0\)
\(\Leftrightarrow 2\leq A\leq 4\Rightarrow \left\{\begin{matrix} A_{\min}=2\\ A_{\max}=4\end{matrix}\right.\)
\(A_{\min}=1\Leftrightarrow x=1\)
\(A_{\max}=4\Leftrightarrow x=-1\)
+)\(A=\dfrac{3x^2-2x+3}{x^2+1}\)
\(A-2=\dfrac{3x^2-2x+3}{x^2+1}-\dfrac{2\left(x^2+1\right)}{x^2+1}\)
\(A-2=\dfrac{x^2-2x+1}{x^2+1}\)
\(A-2=\dfrac{\left(x-1\right)^2}{x^2+1}\ge0\)
\(\Rightarrow A\ge2\)
\(\Rightarrow MINA=2\Leftrightarrow x=1\)
+)\(A=\dfrac{3x^2-2x+3}{x^2+1}\)
\(A-4=\dfrac{3x^2-2x+3}{x^2+1}-\dfrac{4\left(x^2+1\right)}{x^2+1}\)
\(A-4=\dfrac{-x^2-2x-1}{x^2+1}\)
\(A-4=\dfrac{-\left(x+1\right)^2}{x^2+1}\le0\)
\(\Rightarrow A\le4\)
\(\Rightarrow MAXA=4\Leftrightarrow x=-1\)
