a).

\(x^5+x+1=\left(x^5+x^4+x^3\right)-\left(x^4+x^3+x^2\right)+\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left(x^3-x^2\right)\)

b).\(x^8+x^7+1=\left(x^8+x^7+x^6\right)-\left(x^6+x^5+x^4\right)+\left(x^5+x^4+x^3\right)-\left(x^3+x^2+x\right)+\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)

d).

\(x^7+x^5+1=\left(x^7+x^6+x^5\right)-\left(x^6+x^5+x^4\right)+\left(x^5+x^4+x^3\right)-\left(x^3+x^2+x\right)+\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left(x^5-x^4+x^3-x+1\right)\)

e).

\(x^8+x^4+1=x^8+2x^4+1-x^4\\ =\left(x^4+1\right)^2-\left(x^2\right)^2\\ =\left(x^4+x^2+1\right)\left(x^4-x^2+1\right)\\ =\left(x^4-x^2+1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)\)

Phân tích đa thức thành nhân tử

a, 5x-10\(x^2\)

b, \(\dfrac{1}{2}x\left(x^2-4\right)+4\left(y+2\right)\)

c, \(x^4-y^6\)

d, \(x^3+y\left(1-3x^2\right)+z\left(3y^2-1\right)-y^3\)

e, \(x^3-4x^2+4x-1\)

f, \(x^2+2xy-8y^2+2xz+14yz+3z^2\)

g, \(x^4+6x^3-12x^2-8x\)

h, \(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2\)

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

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

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

g) \(x^4+6x^3-12x^2-8x\)

= \(x\left(x^3-2x^2+8x^2-16x+4x-8\right)\)

= \(x\left[x^2\left(x-2\right)+8x\left(x-2\right)+4\left(x-2\right)\right]\)

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

h) \(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2\) (*)

Đặt \(x^2+4x+8=a\) => (*) trở thành:

\(a^2+3ax+2x^2\) = \(a^2+ãx+2ax+x^2\)

= \(a\left(a+x\right)+2x\left(a+x\right)\)

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

Thay \(a=x^2+4x+8\) vào (1) ta được:

\(\left(x^2+5x+8\right)\left(x^2+6x+8\right)\)

=\(\left(x^2+5x+8\right)\left(x^2+2x+4x+8\right)\)

= \(\left(x^2+5x+8\right)\left[x\left(x+2\right)+4\left(x+2\right)\right]\)

= \(\left(x+2\right)\left(x+4\right)\left(x^2+5x+8\right)\)

P/s: Còn câu f đang suy nghĩ!