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

mk ghi đáp án, còn lại bạn tự biến đổi

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

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

c) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)

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

e) \(x^4-7x^3+14x^2-7x+1=\left(x^2-4x+1\right)\left(x^2-3x+1\right)\)

mk làm chi tiết theo yêu của của người hỏi đề:

a) \(2x^3-x^2+5x+3\)

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

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

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

b)  \(x^3+5x^2+8x+4\)

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

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

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

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

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)

Sorry mk ghi nhầm 

C= (x²-1)(x²+2x+4)=(x+1)(x-1)(x²+2x+4)

thì m cứ tách bọn nó ra. tách cho bao giờ ko tách dc nữa thì thôi , nếu tách mãi tách mãi mà vẫn ko dc thì mày kết luận 1 câu là đề sai thế thôi

\(4x^4+4x^3+5x^2+6x+1\)

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

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

\(=\left(x+1\right).\left(4x+5x+1\right)\)

p/s: tớ nghĩ sai đề nên đổi ạ :))

b)=4a²-8ab+4b²-a²-2ab-b²

=3a²-6ab+3b²

=3(a²-2ab+b²)

=3(a-b)²

c, 4 - y^2 - x^2 + 2xy

= 4-(x^2 - 2xy + y^2) 

= 4-(x - y)^2 

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

d, x^2 - 2x - 80

= x^2 + 8x - 10x - 80

= x(x + 8) - 10(x + 8)

= (x - 10(x + 8)

Câu 2 nha

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

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

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

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

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

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

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

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

xog a

tốt

Bài làm

a) 3x² - 6x² + 3x

= -3x² + 3x

= 3x( 1 - x )

b) 3x² + 5x - 3xy - 5y

= ( 3x² - 3xy ) + ( 5x - 5y )

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

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

c) x³ + 2x² + x

= x( x² + 2x + 1 )

= x( x² + 2.x.1 + 1² )

= x( x + 1 )²

d) xy + y² - x - y

= ( xy - x ) + ( y² - y )

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

= ( y - 1 )( x +  y )

# Học tốt #

a)  \(x^4+2x^3+5x^2+4x-12\)

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

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

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

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