Question

[༺ ๖ۣۜPhạm ✌Tuấn ✌Kiệτ ༻]

Tìm GTNN :

\(D=x^4-2x^3+3x^2-2x+1\)

\(E=x^4-6x^3+10x-6x+9\)

Answer 1

[ngonhuminh]

\(D=\left(x^4-2x^3+x^2\right)+\left(2x^2-2x+1\right)\)

\(D=\left(x^2-x\right)^2+2\left(x^2-x\right)+1=\left(x^2-x+1\right)^2\)

\(D=\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]^2\)

\(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\in R\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

\(\Rightarrow D\ge\left(\dfrac{3}{4}\right)^2=\dfrac{9}{16}\)

đẳng thúc khi x=1/2

Answer 2

[ngonhuminh]

{logic 10x-->10x^2}

\(E=x^4-6x^3+10x^2-6x+9\)

\(E=\left(x^4-3x+9x^2\right)+\left(x^2-6x+9\right)\)

\(E=\left(x^2-3x\right)^2+\left(x-3\right)^2=\left[x^2\left(x-3\right)^2\right]+\left(x-3\right)^2\)

\(E=\left(x-3\right)^2\left(x^2+1\right)\)

\(\left\{{}\begin{matrix}\left(x-3\right)^2\ge0\\\left(x^2+1\right)\ge1\end{matrix}\right.\) \(\Rightarrow E\ge0\) đẳng thức khi x=3