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)

Bài làm:

1) Ta có: \(2x^2+5xy+2y^2\)

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

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

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

2) Ta có: \(2x^2+2xy-4y^2\)

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

\(=2x\left(x-y\right)+4y\left(x-y\right)\)

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

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

a)45+x³-5x²-9x

(45-9x)+(x³-5x²)

9(5-x)+x²(x-5)

(9-x²)(5-x)

(3-x)(3+x)(5-x)

b)x⁴-2x³-2x²-2x-3

x³(x-2)-2x(x-2)-3

(x-2)(x³-2x)-3

x

a, \(x^3-2x-4\) b, \(x^2+4x+3\) nhá

Nghịch xíu :v

a, \(x^3-2x-4\)

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

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

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

b, \(x^2+4x+3\)

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

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

Chúc bạn học tốt!!!

=x³(x+2)-13x²+12x-26x+24

=x³(x+2)-x(13x-12)-2(13x-12)

=x³(x+2)-(13x-12)(x+2)

=(x+2)(x³-x-12x+12)

(x+2)[(x²-1)-12(x-1)]

=(x+2)[x(x-1)(x+1)-12(x-1)]

=(x+2)(x-1)[x(x+1)-12]

=(x+2)(x-1)(x²+x-12)

=(x+2)(x-1)(x²-3x+4x-12)

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

=(x+2)(x-1)(x-3)(x+4)

trong bài làm của mk có hàng k có dấu "=" chỗ đó có dâu"=" nha!

x²⁺y²-x²y²+xy-x-y=x²-x²y²+y²-y-x+xy

                            =x(1-y²)+y(y-1)-x(1-y)

                            =x²(y-1)(y+1)+y(y-1)+x(y-1)

                           =-x²(y-1)(y+1)+y(y-1)+x(y-1)

                           =(y-1)(-x²(y+1)+y+x)

f)    x⁴+2x²-4x-4=(x³.x+x³.2)-(2x.2+2.2)

                          =x³(x+2)-2(x+2)

                            =(x³-2)(x+2)

a , 3x² + 3y² - 6xy - 12

= 3 ( x² + y² - 2xy - 4 )

= 3 ( x - y )² - 2²

^{= 3 ( x - y + 2 ) ( x - y - 2 )}

\(x^{40}+2x^{20}+9\)

\(=\left(x^{40}+6x^{20}+9\right)-4x^{20}\)

\(=\left(x^{20}+3\right)^2-\left(2x^{10}\right)^2\)

\(=\left(x^{20}+2x^{10}+3\right)\left(x^{20}-2x^{10}+3\right)\)

x.x-2.x-1-y.y

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

Bài giải:

a) x³ + 2x²y + xy²– 9x = x(x² +2xy + y² – 9)

= x[(x² + 2xy + y²) – 9]

= x[(x + y)² – 3²]

= x(x + y – 3)(x + y + 3)

b) 2x – 2y – x² + 2xy – y² = (2x – 2y) – (x² – 2xy + y²)

= 2(x – y) – (x – y)²

= (x – y)[2 – (x – y)]

= (x – y)(2 – x + y)

c) x⁴ – 2x² = x²(x² – (√2)²) = x²(x - √2)(x + √2).

a) x³ + 2x²y + xy²– 9x = x(x² +2xy + y² – 9)

= x[(x² + 2xy + y²) – 9]

= x[(x + y)² – 3²]

= x(x + y – 3)(x + y + 3)

b) 2x – 2y – x² + 2xy – y² = (2x – 2y) – (x² – 2xy + y²)

= 2(x – y) – (x – y)²

= (x – y)[2 – (x – y)]

= (x – y)(2 – x + y)

c) x⁴ – 2x² = x²(x² – (√2)²) = x²(x - √2)(x + √2).