Bài 9: Phân tích đa thức thành nhân tử bằng cách phối hợp nhiều phương pháp

a)

\(x^3-7x-6=x^3-x-6x-6\)

\(=x(x^2-1)-6(x+1)\)

\(=x(x-1)(x+1)-6(x+1)=(x+1)[x(x-1)-6]\)

\(=(x+1)(x^2-x-6)=(x+1)[x^2-3x+2x-6]\)

\(=(x+1)[x(x-3)+2(x-3)]=(x+1)(x+2)(x-3)\)

b) \(x^3-6x^2+8x\)

\(=x(x^2-6x+8)\)

\(=x(x^2-4x-2x+8)\)

\(=x[x(x-4)-2(x-4)]=x(x-2)(x-4)\)

c) \(x^4+2x^3-16x^2-2x+15\)

\(=(x^4+2x^3-x^2-2x)-15x^2+15\)

\(=[(x^4-x^2)+(2x^3-2x)]-15(x^2-1)\)

\(=[x^2(x^2-1)+2x(x^2-1)]-15(x^2-1)\)

\(=(x^2-1)(x^2+2x)-15(x^2-1)=(x^2-1)(x^2+2x-15)\)

\(=(x^2-1)(x^2-3x+5x-15)=(x^2-1)[x(x-3)+5(x-3)]\)

\(=(x^2-1)(x+5)(x-3)=(x-1)(x+1)(x+5)(x-3)\)

d)

\(x^3-11x^2+30x=x(x^2-11x+30)\)

\(=x(x^2-5x-6x+30)\)

\(=x[x(x-5)-6(x-5)]=x(x-6)(x-5)\)

a) x³ -7x -6

=x³ -x-6x-6

= x(x² -1)-6(x-1)

= x(x-1)(x+1)-6(x-1)

=(x-1)(x² +x+6)

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

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

\(A=\left(a^2-a\right)^2+\left(\sqrt{2}a-\sqrt{2}\right)^2+3\)

Do \(\left(a^2-a\right)^2+\left(\sqrt{2}a-\sqrt{2}\right)^2\ge0\forall a\)

Nên \(\left(a^2-a\right)^2+\left(\sqrt{2}a-\sqrt{2}\right)^2+3\ge3\forall a\)

Dấy "=" xả ra khi a = 1

Vậy Min A = 3 khi a = 1