Câu hỏi của Nam Lee

Question

Tìm  GTNN của :

D = x4 + 2x2 + 3x2 + 2x + 1

Nguyễn Việt Lâm · Accepted Answer

Sửa đề: $D=x^4+2x^3+3x^2+2x+1$

$=x^4+x^3+x^2+x^3+x^2+x+x^2+x+1$

$=x^2\left(x^2+x+1\right)+x\left(x^2+x+1\right)+x^2+x+1$

$=\left(x^2+x+1\right)^2$

$=\left[\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\right]^2\ge\left(\frac{3}{4}\right)^2=\frac{9}{16}$

$D_{min}=\frac{9}{16}$ khi $x=-\frac{1}{2}$Câu trả lời: Sửa đề: $D=x^4+2x^3+3x^2+2x+1$

$=x^4+x^3+x^2+x^3+x^2+x+x^2+x+1$

$=x^2\left(x^2+x+1\right)+x\left(x^2+x+1\right)+x^2+x+1$

$=\left(x^2+x+1\right)^2$

$=\left[\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\right]^2\ge\left(\frac{3}{4}\right)^2=\frac{9}{16}$

$D_{min}=\frac{9}{16}$ khi $x=-\frac{1}{2}$

Violympic toán 8