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)

a: Ta có: \(10x^4-27x^3y-110x^2y^2-27xy^3+10y^4\)

\(=10x^4+20x^2y^2+10y^4-27xy\left(x^2+y^2\right)-130x^2y^2\)

\(=10\left(x^2+y^2\right)^2-27xy\left(x^2+y^2\right)-130x^2y^2\)

\(=10\left(x^2+y^2\right)^2-52xy\left(x^2+y^2\right)+25xy\left(x^2+y^2\right)-130x^2y^2\)

\(=2\left(x^2+y^2\right)\left(5x^2+5y^2-26xy\right)+5xy\left(5x^2+5y^2-26xy\right)\)

\(=\left(5x^2-26xy+5y^2\right)\left(2x^2+5xy+2y^2\right)\)

\(=\left(5x^2-25xy-xy+5y^2\right)\left(2x^2+4xy+xy+2y^2\right)\)

\(=\left\lbrack5x\left(x-5y\right)-y\left(x-5y\right)\right\rbrack\left\lbrack2x\left(x+2y\right)+y\left(x+2y\right)\right\rbrack\)

=(5x-y)(x-5y)(2x+y)(x+2y)

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

\(=x^5+x^4-5x^4-5x^3+8x^3+8x^2-5x^2-5x+x+1\)

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

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

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

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

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

tách ra như bth ấy

Câu 1 :

a) \(x^3-5x^2-14x\)

\(=x^3-7x^2+2x^2-14x\)

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

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

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

b) \(a^4+a^2+1\)

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

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

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

c) \(x^4+64\)

\(=\left(x^2\right)^2+2\cdot x^2\cdot8+8^2-2\cdot x^2\cdot8\)

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

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

Câu 2 :

a) \(\left(a-b\right)^2=a^2-2ab+b^2\)

Ta có : \(\left(a+b\right)^2=a^2+2ab+b^2\)

\(\Rightarrow a^2+b^2=\left(a+b\right)^2-2ab=7^2-2\cdot14=25\)

\(\Rightarrow\left(a-b\right)^2=25-2\cdot12=1\)

b) tương tự

\(a^4+8a^3+14a^2-8a-15\)

\(=a^4+8a^3+15a^2-a^2-8a-15\)

\(=a^2\left(a^2+8a+15\right)-\left(a^2+8a+15\right)\)

\(=\left(a^2+8a+15\right)\left(a^2-1\right)\)

\(=\left(a+3\right)\left(a+5\right)\left(a-1\right)\left(a+1\right)\)

a, \(x^4-x^3-x^3+x^2-x^2+x+x-1\)\(1\)

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

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

b, \(\left(ab-1\right)^2+\left(a+b\right)^2\)

=\(a^2b^2-2ab+1+a^2+2ab+b^2\)

=\(a^2b^2+a^2+b^2+1\)

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

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

c,\(x^4+2x^3+2x^2+2x+1\)

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

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

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

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

Trả lời:

1) sửa đề:  \(x^4+x^3-4x-4=x^3\left(x+1\right)-4\left(x+1\right)=\left(x+1\right)\left(x^3-4\right)\)

2) \(x^2-\left(a+b\right)x+ab=x^2-ax-bx+ab=\left(x^2-ax\right)-\left(bx-ab\right)\)

\(=x\left(x-a\right)-b\left(x-a\right)=\left(x-a\right)\left(a-b\right)\)

3)  \(5xy^3-2xyz-15y^2+6z=\left(5xy^3-15y^2\right)-\left(2xyz-6z\right)\)

\(=5y^2\left(xy-3\right)-2z\left(xy-3\right)=\left(xy-3\right)\left(5y^2-2z\right)\)

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

\(\Leftrightarrow\left(x+1-x+1\right)\left(x+1+x-1\right)\)

\(\Leftrightarrow2.2x=4x\)

p/s tham khảo nha

\(a^2-b^2-a+b\)

\(\Leftrightarrow\left(a-b\right)\left(a+b\right)-\left(a-b\right)\)

\(\Leftrightarrow\left(a-b\right)\left(a+b-1\right)\)

p/s tham khảo

a) x² – 4x + 3 = x² – x - 3x + 3

= x(x - 1) - 3(x - 1) = (x -1)(x - 3)

b) x² + 5x + 4 = x² + 4x + x + 4

= x(x + 4) + (x + 4)

= (x + 4)(x + 1)

c) x² – x – 6 = x² +2x – 3x – 6

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

= (x + 2)(x - 3)

d) x⁴+ 4 = x⁴ + 4x² + 4 – 4x²

= (x² + 2)² – (2x)²

= (x² + 2 – 2x)(x² + 2 + 2x)

Bài giải:

a) x² – 4x + 3 = x² – x - 3x + 3

= x(x - 1) - 3(x - 1) = (x -1)(x - 3)

b) x² + 5x + 4 = x² + 4x + x + 4

= x(x + 4) + (x + 4)

= (x + 4)(x + 1)

c) x² – x – 6 = x² +2x – 3x – 6

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

= (x + 2)(x - 3)

d) x⁴+ 4 = x⁴ + 4x² + 4 – 4x²

= (x² + 2)² – (2x)²

= (x² + 2 – 2x)(x² + 2 + 2x)