Question

Chứng minh x^4 -x +1 > 0

Answer 1

Lời giải:

Ta có:

$x^4-x+1=(x^4-x^2+\frac{1}{4})+(x^2-x+\frac{1}{4})+\frac{1}{2}$

$=(x^2-\frac{1}{2})^2+(x-\frac{1}{2})^2+\frac{1}{2}$

$\geq 0+0+\frac{1}{2}>0$

Ta có đpcm.

Answer 2

Ta có: \(x^4-x+1\)

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

\(=\left(x^2-\dfrac{1}{2}\right)^2+\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{2}>0\forall x\)