Câu hỏi của Nguyễn Chí Thành

Question

Chứng minh rằng $x^4+4x^3+6x^2+4x+1\ge0,\forall x\in R$

Nguyễn Việt Lâm · Accepted Answer

$x^4+4x^3+6x^2+4x+1$

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

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

$=\left(x^2+2x+1\right)\left(x^2+2x+1\right)=\left(x+1\right)^4\ge0;\forall x\in R$Câu trả lời: $x^4+4x^3+6x^2+4x+1$

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

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

$=\left(x^2+2x+1\right)\left(x^2+2x+1\right)=\left(x+1\right)^4\ge0;\forall x\in R$