\(P\left(x\right)=\dfrac{1}{2}x^3-\dfrac{1}{2}x^4+\dfrac{1}{2}x^2+\dfrac{1}{2}x^4-x^2=-\dfrac{1}{2}x^3+\dfrac{1}{2}x^2=-\dfrac{1}{2}x^2\left(x-1\right)\)

Vì x(x-1) chia hết cho 2 với mọi số nguyên x 

nên P(x) luôn là số nguyên nếu x nguyên

a) \(A=x^2-2x+2=\left(x-1\right)^2+1>0\forall x\inℝ\)

b) \(x-x^2-3=-\left(x^2-x+3\right)\)

\(=-\left(x^2-x+\frac{1}{4}+\frac{11}{4}\right)\)

\(=-\left[\left(x-\frac{1}{2}\right)^2+\frac{11}{4}\right]\)

\(=-\left[\left(x-\frac{1}{2}\right)^2\right]-\frac{11}{4}\le\frac{-11}{4}< 0\forall x\inℝ\)

x²-2x+2=(x²-2x+1)+1=( x-1)²+1

Mà (x-1)²≥0 với mọi x

=> (x-1)²+1>0 với mọi x

=> x²-2x+2>0 với mọi x

P = \(x^4-2x^3+2x^2-2x+1\)

P = \(x^4-x^3-x^3+x^2+x^2-x-x+1\)

P = \(x^3\left(x-1\right)-x^2\left(x-1\right)+x\left(x-1\right)-\left(x-1\right)\)

P = \(\left(x-1\right)\left(x^3-x^2+x-1\right)\)

P = \(\left(x-1\right)\left[x^2\left(x-1\right)+\left(x-1\right)\right]\)

P = \(\left(x-1\right)\left(x-1\right)\left(x^2+1\right)\)

P = \(\left(x-1\right)^2\left(x^2+1\right)\) \(\ge\forall x\) ( đpcm )

Chúc bạn học tốt :))

 = (x2-x+1)(x2+3x+10)+10 = P

x²-x+1=(x-\(\frac{1}{2}\))²+\(\frac{3}{4}\)>0

x²+3x+10=(x+\(\frac{3}{2}\))²+\(\frac{31}{4}\)>0

vây P>0

\(2x^4+1\ge2x^3+x^2\)

\(\Leftrightarrow2x^4-2x^3-x^2+1\ge0\)

\(\Leftrightarrow\left(x^4-2x^3+x^2\right)+\left(x^4-2x^2+1\right)\ge0\)

\(\Leftrightarrow\left(x^2-x\right)^2+\left(x^2-1\right)^2\ge0\) đúng

Đặt A=x^4-x^3+3x^2-2x+2

=(x^4+3x^2+2)-(x^3+2x)

=(x^4+x^2+2x^2+2)-x(x^2+2)

=(x^2+1)(x^2+2)-x(x^2+2)

=(x^2+2)(x^2-x+1)

Ta có x^2+2>=2>0;

x^2-x+1=(x^2-x+1/4)+3/4 =(x-1/2)^2+3/4>=3/4>0 

=> A>0