a) \(B=x^2-3x+5\Leftrightarrow x^2-2.x.\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2-\left(\dfrac{3}{2}\right)^2+5\)
\(B=\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\forall x\)
\(\Rightarrow minB=\dfrac{11}{4}\) khi \(\left(x-\dfrac{3}{2}\right)^2=0\Leftrightarrow x-\dfrac{3}{2}=0\Leftrightarrow x=\dfrac{3}{2}\)
vậy GTNN của B là \(\dfrac{11}{4}\)khi \(x=\dfrac{3}{2}\)
b) \(C=x^2-x+6\Leftrightarrow x^2-2.x\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{2}\right)^2+6\)
\(C=\left(x-\dfrac{1}{2}\right)^2+\dfrac{23}{4}\ge\dfrac{23}{4}\forall x\)
\(\Rightarrow minC=\dfrac{23}{4}\) khi \(\left(x-\dfrac{1}{2}\right)^2=0\Leftrightarrow x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}\)
vậy GTNN của C là \(\dfrac{23}{4}\) khi \(x=\dfrac{1}{2}\)
a, Theo bài ra ta có:
\(B=x^2-\dfrac{3}{2}.2.x+\left(\dfrac{3}{2}\right)^2+\dfrac{11}{4}\)
\(B=\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{4}\)
\(=>B\ge\dfrac{11}{4}\)
\(=>MinB=\dfrac{11}{4}\Leftrightarrow a=\dfrac{3}{2}\)
b,Theo bài ra ta có:
\(C=x^2-\dfrac{1}{2}.2.x+\left(\dfrac{1}{2}\right)^2+\dfrac{23}{4}\)
\(C=\left(x-\dfrac{1}{2}\right)^2+\dfrac{23}{4}\)
\(=>C\ge\dfrac{23}{4}\)
\(=>MinC=\dfrac{23}{4}\Leftrightarrow x=\dfrac{1}{2}\)
CHÚC BẠN HỌC TỐT...........
