Bài 2:

a: \(x^4-x^3+6x^2-x+a\)

\(=x^4-x^3+5x^2+x^2-x+5+a-5\)

\(=\left(x^2-x+5\right)\left(x^2+1\right)+a-5\)

TA có: \(x^4-x^3+6x^2-x+a\) ⋮\(x^2-x+5\)

=>a-5=0

=>a=5

b: \(x^4+a\cdot x^2+b\)

\(=x^4-x^3+x^2+x^3-x^2+x+a\cdot x^2-a\cdot x+a\) +x(a-1)+b-a

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

Để \(x^4+a\cdot x^2+b\) ⋮\(x^2-x+1\) thì a-1=0 và b-a=0

=>a=b=1

4a^2 + 4a + 2 = ( 2a)^2 + 2.2a.1 + 1 + 1

                    = ( 2a + 1)^2 + 1

Vì ( 2a + 1)^2 lớn hơn bằng 0 => ( 2a + 1)^2 + 1 lớn hơn bằng 0 + 1 = 1

Vậy Min = 1 khi 2a +1 =  => a = - 1/ 2

\(A=a^4-2a^3+3a^2-4a+5\)

\(A=a^4-2a^3+a^2+2a^2-4a+2+3\)

\(A=\left(a^4-2a^3+a^2\right)+\left(2a^2-4a+2\right)+3\)

\(A=\left(a^2-a\right)^2+2\left(a-1\right)^2+3\)

Ta có: \(\left(a^2+a\right)^2\ge0\) với mọi x

và: \(2\left(a-1\right)^2\ge0\)

Suy ra: \(A\ge3\)

Vậy min A = 3 khi a = 1

a: Ta có: \(x^4-2x^3+2x-1\)

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

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

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

b: Ta có: \(-a^4+a^3+2a^3+2a^2\)

\(=-a^2\left(a^2-a-2a-2\right)\)

c: Ta có: \(x^4+x^3+2x^2+x+1\)

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

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