\(A=x^2+x+1\)
\(A=\left(x^2+2x+1\right)-x\)
\(A=\left(x+1\right)^2-x\)
Ta có: \(\left(x+1\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+1\right)^2-x\ge-x\)
\(\Rightarrow Min_A=-x\) khi \(x+1=0\Leftrightarrow x=-1\)
làm lại nhé.sorry
A=(x^2+x+1)^2
tìm MIN
bó tay!!
\(A=\left(x^2+x+1\right)^2\)
Ta có: \(\left(x^2+x+1\right)^2\ge0\forall x\)
\(\Rightarrow Min_A=0\)
(Làm đại không chắc)
\(B=x^2+x+1\)
\(=x^2+2\cdot\dfrac{1}{2}\cdot x+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\) \(\ge\dfrac{3}{4}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x+\dfrac{1}{2}=0\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
Do đó : \(A=B^2\ge\dfrac{9}{16}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=-\dfrac{1}{2}\)
