\(A=x^2+2x+6=(x^2+2x+1)+5=(x+1)^2+5\)
Vì \((x+1)^2\geq 0, \forall x\in\mathbb{R}\Rightarrow A\geq 0+5=5\)
Vậy GTNN của $A$ là $5$ khi $(x+1)^2=0$ hay $x=-1$
--------------
\(B=x^2-6x+15=(x^2-2.3x+3^2)+6=(x-3)^2+6\)
Vì \((x-3)^2\geq 0, \forall x\in\mathbb{R}\Rightarrow B\geq 0+6=6\)
Vậy GTNN của $B$ là $6$ khi $x=3$
---------------
\(C=x^2-5x+3=x^2-2.\frac{5}{2}x+(\frac{5}{2})^2-\frac{13}{4}=(x-\frac{5}{2})^2-\frac{13}{4}\)
Vì \((x-\frac{5}{2})^2\geq 0, \forall x\in\mathbb{R}\Rightarrow C\geq 0-\frac{13}{4}=\frac{-13}{4}\)
Vậy \(C_{\min}=\frac{-13}{4}\Leftrightarrow x=\frac{5}{2}\)
\(D=2x^2-7x+1=2(x^2-\frac{7}{2}x)+1\)
\(=2[x^2-2.\frac{7}{4}x+(\frac{7}{4})^2]-\frac{41}{8}\)
\(=2(x-\frac{7}{4})^2-\frac{41}{8}\)
Vì \((x-\frac{7}{4})^2\geq 0, \forall x\in\mathbb{R}\Rightarrow D\geq 2.0-\frac{41}{8}=-\frac{41}{8}\)
Vậy \(D_{\min}=-\frac{41}{8}\Leftrightarrow x=\frac{7}{4}\)
--------------------
\(E=3x^2+2x=3(x^2+\frac{2}{3})=3[x^2+2.\frac{1}{3}x+(\frac{1}{3})^2]-\frac{1}{3}\)
\(=3(x+\frac{1}{3})^2-\frac{1}{3}\)
Vì \((x+\frac{1}{3})^2\geq 0, \forall x\in\mathbb{R}\Rightarrow E\geq 3. 0-\frac{1}{3}=\frac{-1}{3}\)
Vậy \(E_{\min}=\frac{-1}{3}\Leftrightarrow x=\frac{-1}{3}\)
