b) Ta có:
\(x^2-5x+3=x^2-2.x.\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2-\dfrac{13}{4}=\left(x-\dfrac{5}{2}\right)^2-\dfrac{13}{4}\ge-\dfrac{13}{4}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x-\dfrac{5}{2}\right)^2=0\Leftrightarrow x=\dfrac{5}{2}\)
Vậy, Minx2 - 5x + 3 = \(-\dfrac{13}{4}\) \(\Leftrightarrow x=\dfrac{5}{2}\)
a) Ta có:
\(x^2+6x+15=x^2+2.x.3+3^2+6=\left(x+3\right)^2+6\ge6\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x+3\right)^2=0\Leftrightarrow x=-3\)
Vậy Minx2 + 6x + 15 = 6 \(\Leftrightarrow x=-3\)
\(a.x^2+6x+15=x^2+6x+9+6=\left(x+3\right)^2+6\)
Ta có: \(\left(x+3\right)^2\ge0\forall x\Rightarrow\left(x+3\right)^2+6\ge6\forall x\)
Dấu "=" xảy ra <=> \(\left(x+3\right)^2=0\Leftrightarrow x=-3\)
\(\Rightarrow Min=6\Leftrightarrow x=-3\)
\(c.C=x^2+x+1=x^2+x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Vì \(\left(x+\dfrac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)
\(\Rightarrow Min_C=\dfrac{3}{4}\)
Dấu "=" xảy ra khi \(\left(x+\dfrac{1}{2}\right)^2=0\Leftrightarrow x=\dfrac{-1}{2}\)
