mk viết đáp án, ko biết biến đổi ib mk

a)  \(x^3+3x^2y-9xy^2+5y^3=\left(x+5y\right)\left(x-y\right)^2\)

b)    \(x^4+x^3+6x^2+5x+5=\left(x^2+5\right)\left(x^2+x+1\right)\)

c)   \(x^4-2x^3-12x^2+12x+36=\left(x^2-6\right)\left(x^2-2x-6\right)\)

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

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)

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) x² – 4 + (x – 2)²

= (x² – 2²) + (x – 2)² = (x – 2)(x + 2) + (x – 2)²

= (x – 2) [(x + 2) + (x – 2)]

= (x – 2)(x + 2 + x – 2)

= 2x(x – 2)

b) 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)

c) x³ – 4x² – 12x + 27

= (x³ + 27) – 4x(x + 3)

= (x + 3)(x² – 3x + 9) – 4x(x + 3)

= (x + 3)(x² – 3x + 9 – 4x)

= (x + 3)(x² – 7x + 9)

câu a nè = (4x-1)(2x-3) 

câu f = (x+y+z) ( x^ 2 + y^2 + z^2 +xy + yz + zx)

Có câu nào khó hơn không bạn

\(e,-5x+x^2-14\)

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

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

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

\(f,x^3+8+6x\left(x+2\right)\)

\(=\left(x+2\right)\left(x^2+2x+4\right)+6x\left(x+2\right)\)

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

\(g,15x^2-7xy-2y^2\)

\(=15x^2+3xy-10xy-2y^2\)

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

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

\(h,3x^2-16x+5\)

\(=3x^2-x-15x+5\)

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

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

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

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

\(b,4x^2-9y^2+4x-6y\)

\(=4x^2+4x+1-\left(9y^2+6y+1\right)\)

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

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

\(c,-x^2+5x+2xy-5y-y^2\)

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

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

\(=\left(x-y\right)\left(y-x+5\right)\)

\(d,x^2+4x-12\)

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

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

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