${{x}^{8}}-{{x}^{5}}+{{x}^{2}}-x+1$
Xét $x \le 0$ thì ${{x}^{8}} \ge 0$; $-{{x}^{5}} \ge 0$; ${{x}^{2}} \ge 0$; $-x \ge 0$
Khi đó $P={{x}^{8}}-{{x}^{5}}+{{x}^{2}}-x+1 \ge 1 > 0$
Xét $0 < x < 1$
Ta có: $P={{x}^{8}}+{{x}^{2}}\left( 1-{{x}^{3}} \right)+\left( 1-x \right)$
Vì $0 < x < 1$ nên $\left( 1-{{x}^{3}} \right) > 0$; $\left( 1-x \right) > 0$
Khi đó $P={{x}^{8}}+{{x}^{2}}\left( 1-{{x}^{3}} \right)+\left( 1-x \right) > 0$
Xét $x \ge 1$
Ta có: $P={{x}^{5}}\left( {{x}^{3}}-1 \right)+x\left( x-1 \right)+1$
Vì $x \ge 1$ nên $\left( {{x}^{3}}-1 \right) \ge 0$; $\left( x-1 \right) \ge 0$
Khi đó $P={{x}^{5}}\left( {{x}^{3}}-1 \right)+x\left( x-1 \right)+1 \ge 1 > 0$
Vậy P(x) luôn dương với mọi giá trị x