c/ Ta có:

\(x^2-3xy+x-3y\)

\(=x^2+x-3xy-3y\)

\(=x\left(x+1\right)-3y\left(x+1\right)\)

\(=\left(x+1\right)\left(x-3y\right)\)

d/ Ta có:

\(x^3-x^2-5x+125\)

\(=x^3+5x^2-6x^2-30x+25x+125\)

\(=x^2\left(x+5\right)-6x\left(x+5\right)+25\left(x+5\right)\)

\(=\left(x+5\right)\left(x^2-6x+25\right)\)

\(x^2-3xy+x-3y\)

\(=x\left(x-3y\right)+\left(x-3y\right)\)

\(=\left(x+1\right)\left(x-3y\right)\)

\(x^3-x^2-5x+125\) k có nghiệm

\(C=x^3+5x^2+8x+4\)

   \(=x^3+x^2+4x^2+4x+4x+4\)

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

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

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

\(D=x^3-x^2-4\)

    \(=x^3-2x^2+x^2-2x+2x-4\)

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

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

Chúc bạn học tốt.

Dùng hằng đẳng thức là xong

a, \(\left(x+y\right)^3-x^3-y^3=x^3+3x^2y+3xy^2+y^3-x^3-y^3\)

\(=3x^2y+3xy^2=3xy\left(x+y\right)\)

b,  \(x^2+6xy+9y^2=\left(x+3y\right)^2\)

       \(x^3+5x^2+3x-9\)

\(=x^3-x^2+6x^2-6x+9x-9\)

\(=x^2\left(x-1\right)+6x\left(x-1\right)+9\left(x-1\right)\)

\(=\left(x-1\right)\left(x^2+6x+9\right)=\left(x-1\right)\left(x+3\right)^2\)

       \(x^{16}+x^8-2\)

\(=\left(x^{16}-1\right)+\left(x^8-1\right)\)

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

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

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

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

\(c,x^3+5x^2+3x-9\)

\(=x^3+6x^2+9-x^2-6x-9\)

\(=x\left(x^2+6x^2+9\right)-\left(x^2+6x^2+9\right)\)

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

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

\(d,x^{16}+x^8-2\)

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

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

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

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

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

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

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

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

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

\(x^2-y^2+2x+1\)

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

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

\(=\left(x-y+1\right)\left(x+y+1\right)\)

hk tốt

^^

a) \(x^{5} - x^{4} - 2 x^{3} + 2 x^{2} + x - 1\)
Nhóm các hạng tử:

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

Đặt \(t = x^{2}\) thì \(x^{4} - 2 x^{2} + 1 = \left(\right. t - 1 \left.\right)^{2} = \left(\right. x^{2} - 1 \left.\right)^{2}\).
Vậy

\(\boxed{x^{5} - x^{4} - 2 x^{3} + 2 x^{2} + x - 1 = \left(\right. x - 1 \left.\right) \left(\right. x^{2} - 1 \left.\right)^{2} = \left(\right. x - 1 \left.\right)^{3} \left(\right. x + 1 \left.\right)^{2} .}\)

b) \(x^{3} - 5 x^{2} - 14 x\)
Lấy \(x\) chung:

\(x \left(\right. x^{2} - 5 x - 14 \left.\right) = x \left(\right. x - 7 \left.\right) \left(\right. x + 2 \left.\right) .\)

\(\boxed{x^{3} - 5 x^{2} - 14 x = x \left(\right. x - 7 \left.\right) \left(\right. x + 2 \left.\right) .}\)

c) \(2 x^{2} + 2 x y - 4 y^{2}\)
Lấy \(2\) chung: \(2 \left(\right. x^{2} + x y - 2 y^{2} \left.\right)\).
Nhân tử hóa: \(x^{2} + x y - 2 y^{2} = \left(\right. x + 2 y \left.\right) \left(\right. x - y \left.\right)\).
\(\boxed{2 x^{2} + 2 x y - 4 y^{2} = 2 \left(\right. x + 2 y \left.\right) \left(\right. x - y \left.\right) .}\)

d) \(3 x^{2} + 8 x y - 3 y^{2}\)
Thử phân tích:

\(3 x^{2} + 8 x y - 3 y^{2} = \left(\right. 3 x - y \left.\right) \left(\right. x + 3 y \left.\right) .\)

\(\boxed{3 x^{2} + 8 x y - 3 y^{2} = \left(\right. 3 x - y \left.\right) \left(\right. x + 3 y \left.\right) .}\)

e) \(x^{2} - x - x y - 2 y^{2} + 2 y\)
Gộp lại theo \(x\): \(x^{2} + x \left(\right. - 1 - y \left.\right) + \left(\right. - 2 y^{2} + 2 y \left.\right)\).
Định thức là một bình phương → nghiệm \(x = 2 y\) và \(x = 1 - y\).
Vậy

\(\boxed{x^{2} - x - x y - 2 y^{2} + 2 y = \left(\right. x - 2 y \left.\right) \left(\right. x + y - 1 \left.\right) .}\)

f) \(x^{2} + 2 y^{2} - 3 x y + x - 2 y\)
Xem như phương trình bậc hai theo \(x\): nghiệm \(x = 2 y\) và \(x = y - 1\).
Do đó

\(\boxed{x^{2} + 2 y^{2} - 3 x y + x - 2 y = \left(\right. x - 2 y \left.\right) \left(\right. x - y + 1 \left.\right) .}\)

a: \(x^5-x^4-2x^3+2x^2+x-1\)

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

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

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

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

b: \(x^3-5x^2-14x\)

\(=x\left(x^2-5x-14\right)\)

\(=x\left(x^2-7x+2x-14\right)\)

=x[x(x-7)+2(x-7)]

=x(x-7)(x+2)

c: \(2x^2+2xy-4y^2\)

\(=2\left(x^2+xy-2y^2\right)\)

\(=2\left(x^2+2xy-xy-2y^2\right)\)

=2[x(x+2y)-y(x+2y)]

=2(x+2y)(x-y)

d: \(3x^2+8xy-3y^2\)

\(=3x^2+9xy-xy-3y^2\)

=3x(x+3y)-y(x+3y)

=(x+3y)(3x-y)

e: \(x^2-x-xy-2y^2+2y\)

\(=\left(x^2-xy-2y^2\right)-\left(x-2y\right)\)

\(=\left(x^2-2xy+xy-2y^2\right)-\left(x-2y\right)\)

=x(x-2y)+y(x-2y)-(x-2y)

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

f: \(x^2+2y^2-3xy+x-2y\)

\(=x^2-2xy-xy+2y^2+x-2y\)

=x(x-2y)-y(x-2y)+(x-2y)

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

các bạn giúp mình với 

toan lop 8 kho qua.

c, \(2x^2+x-3=x\left(2x-3\right)\)

d, \(6x^2-x-15=x\left(6x-15\right)\)

TK MIK NHA

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

a)4y(x-1)-(x+1)=4y(x-1)-(X-1)

   =(x-1) (4y-1)

b)\(\left(x-3\right)^3\)+3-x=(x-3)^3+(x-3)

=x-3 (x-3)^2

HOK TỐT K NHA