\(A=x^2-x+1\)
=\(\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{3}{4}\)
=\(\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Với mọi x thì \(\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)>=\(\dfrac{3}{4}\)
Hay \(A>=\dfrac{3}{4}\)
Để\(A=\dfrac{3}{4}\) thì
\(\left(x-\dfrac{1}{2}\right)^2=0\)
=>\(x-\dfrac{1}{2}=0\)
=>\(x=\dfrac{1}{2}\)
Vậy...
\(B=x^2-5x-2\)
=\(\left(x^2-5x+\dfrac{25}{4}\right)-\dfrac{33}{4}\)
=\(\left(x-\dfrac{5}{2}\right)^2-\dfrac{33}{4}\)
Với mọi x thì \(\left(x-\dfrac{5}{2}\right)^2-\dfrac{33}{4}\)>=\(-\dfrac{33}{4}\)
Hay\(B>=-\dfrac{33}{4}\)
Để \(B=-\dfrac{33}{4}\)thì ...(giải tìm x)
=>\(x=\dfrac{5}{2}\)
Vậy....
A = \(x^2-x+1\)
=\(x^2-x+\dfrac{1}{4}+\dfrac{3}{4}\)
=\(\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Vì \(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)
\(\Rightarrow A\ge\dfrac{3}{4}\forall x\)
Vậy GTNN của A là \(\dfrac{3}{4}\) khi \(\left(x-\dfrac{1}{2}\right)^2=0\) hay x = \(\dfrac{1}{2}\)
B=\(x^2-5x-2\)
=\(x^2-5x+\dfrac{25}{4}-\dfrac{33}{4}\)
=\(\left(x-\dfrac{5}{2}\right)^2-\dfrac{33}{4}\)
Vì \(\left(x-\dfrac{5}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-\dfrac{5}{2}\right)^2-\dfrac{33}{4}\ge-\dfrac{33}{4}\forall x\)
\(\Rightarrow B\ge-\dfrac{33}{4}\forall x\)
Vậy GTNN của B là \(-\dfrac{33}{4}\) khi \(\left(x-\dfrac{5}{2}\right)^2=0\) hay x=\(\dfrac{5}{2}\)
\(A=x^2-x+1=x^2-x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Dấu "=" xảy ra khi x=1/2
Vậy MinA=3/4
\(B=x^2-5x-2\)
\(=x^2-5x+\dfrac{25}{4}-\dfrac{33}{4}\)
\(=\left(x-\dfrac{5}{2}\right)^2-\dfrac{33}{4}\ge-\dfrac{33}{4}\)
Dấu "=" xảy ra khi x=5/2
Vậy MinB=-33/4
