Em tham khảo:
$x^{6}-x^3+x^2-x+1=(x^3)^2-2.$ $\dfrac{1}{2}x^3+$$\dfrac{1}{4}+x^2-2.x.$$\dfrac{1}{2}+$ $\dfrac{1}{4}+$$\frac{1}{2}$
$=(x^3+\dfrac{1}{2})^{2}+$ $(x^2+\dfrac{1}{2})^{2}+$ $\frac{1}{2}$
Nhận xét
$(x^3+\dfrac{1}{2})^{2}$$\geq0$
$(x^2+\dfrac{1}{2})^{2}$$\geq0$
⇒$(x^3+\dfrac{1}{2})^{2}+$$(x^2+\dfrac{1}{2})^{2}+$$\dfrac{1}{2}>0$
⇒$P>0$
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