1. \(A=x^2-2x+5\)
\(=\left(x^2-2x+1\right)+4\)
\(=\left(x-1\right)^2+4\)
Vì \(\left(x-1\right)^2\ge0\forall x\) nên \(\left(x-1\right)^2+4\ge4\forall x\)
Dấu "=" \(\Leftrightarrow\) x - 1 = 0 \(\Leftrightarrow\) x = 1
Vậy ...
2. \(B=-2x^2-4x+1\)
\(=-2\left(x^2+2x-\dfrac{1}{2}\right)\)
\(=-\left(x^2+2x+1-\dfrac{3}{2}\right)\)
\(=-\left(x+1\right)^2+\dfrac{3}{2}\le\dfrac{3}{2}\forall x\)
Dấu "=" \(\Leftrightarrow\) x + 1 = 0 \(\Leftrightarrow\) x = -1
Vậy ...
3. \(C=\dfrac{3}{-x^2+2x-4}\)
\(=-\dfrac{3}{x^2-2x+4}\)
\(=-\dfrac{3}{\left(x^2-2x+1\right)+3}\)
\(=-\dfrac{3}{\left(x-1\right)^2+3}\)
Vì \(\left(x-1\right)^2\ge0\) nên \(\left(x-1\right)^2+3\ge3\)
\(\Rightarrow-\dfrac{3}{\left(x-1\right)^2+3}\ge-\dfrac{3}{3}=-1\)
Dấu "=" \(\Leftrightarrow\) x - 1 = 0 \(\Leftrightarrow\) x = 1
Vậy ...
