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)

c, \(2x^2+x-3=x\left(2x-3\right)\)

d, \(6x^2-x-15=x\left(6x-15\right)\)

TK MIK NHA

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

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

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

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

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

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

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

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

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

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

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

hk tốt

^^

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

\(x^2-2x-3\)

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

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

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

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

câu a đặt chung x ra là xong
câu b 
x^3 + 3x^2 - 7x^2 - 21x + 9x+ 27 còn lại tự làm nhé

a) x³ - 2x² + x - xy²

= x (x² - 2x + 1 - y²)

= x [(x² - 2x + 1) - y²]

= x [(x - 1)² - y²]

= x [(x - 1) + y] [(x - 1) - y]

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

b) x³ - 4x² - 12x + 27

= (x³ + 27) - (4x² + 12x)

= (x³ + 3³) - 4x (x + 3)

= (x + 3) (x² - 3x + 3²) - 4x (x + 3)

= (x + 3) [(x² - 3x + 9) - 4x]

= (x + 3) (x² - 3x + 9 - 4x)

= (x + 3) (x² - 7x + 9)

#Học tôt!!!

~NTTH~

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!!!

Bạn tách ra thành các dòng để bọn mình dễ nhìn hơn nhé.

a:Sửa đề: \(\left(2x+5\right)^2-\left(x-3\right)^2\)

=(2x+5-x+3)(2x+5+x-3)

=(x+8)(3x+2)

b:Sửa đề: \(25\left(2x-1\right)^2-9\left(x+1\right)^2\)

\(=\left(10x-5\right)^2-\left(3x+3\right)^2\)

=(10x-5-3x-3)(10x-5+3x+3)

=(7x-8)(13x-2)

c: \(1-9x+27x^2-27x^3\)

\(=1^3-3\cdot1^2\cdot3x+3\cdot1\cdot\left(3x\right)^2-\left(3x\right)^3\)

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

d: \(49-a^2+2ab-b^2\)

\(=7^2-\left(a-b\right)^2\)

=(7-a+b)(7+a-b)

e: \(-4x^2-12xy-9y^2+25\)

\(=25-\left(4x^2+12xy+9y^2\right)\)

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

=(5-2x-3y)(5+2x+3y)

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

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

Trả lời:

a) \(x^3+2x=x\left(x^2+2\right)\)

b) \(3x^3-12x^2=3x^2\left(x-4\right)\)

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

b.\(3x^3-12x^2=3x^2\left(x-4\right)\)

Chúc bạn học tốt!