a) $a^2-b^2-4a+4$

$\mathrm{=(a2-4a+4)-b2}$

$\mathrm{=(a-2)2-b2}$

$\mathrm{=(a-2-b)(a-2+b)}$

b) $x^2+2x-3$

$\mathrm{=x2-x+3x-3}$

$\mathrm{=x(x-1)+3(x-1)}$

$\mathrm{=(x+3)(x-1)}$

c) $4x^2{y}^2-{\left(x^2+y^2\right)}^2$

=(2xy-x²-y²)(2xy+x²+y²)

=-(x-y)²(x+y)²

d) $2a^3-54b^3$

=2(a³-27b³)

=2(a-3b)(a2+3ab+9b2)

a) \(\left(4x^2-25\right)^2-9\left(2x-5\right)^2\)

\(=\left(4x^2-25\right)^2-\left(6x-15\right)^2\)

\(=\left(4x^2-25-6x+15\right)\left(4x^2-25+6x-15\right)\)

\(=\left(4x^2-6x-10\right)\left(4x^2+6x-40\right)\)

\(=\left(4x^2+4x-10x-10\right)\left(4x^2+16x-10x-40\right)\)

\(=\left[4x\left(x+1\right)-10\left(x+1\right)\right]\left[4x\left(x+4\right)-10\left(x+4\right)\right]\)

\(=\left(4x-10\right)\left(x+1\right)\left(4x-10\right)\left(x+4\right)\)

\(=\left(4x-10\right)^2\left(x+1\right)\left(x+4\right)\)

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

b) \(a^6-a^4+2a^3+2a^2\)

\(=a^2\left(a^4-a^2+2a+2\right)\)

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

\(=a^2\left[a^3\left(a+1\right)-a^2\left(a+1\right)+2\left(a+1\right)\right]\)

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

Từ điểm B, C vẽ các đường thẳng lần lượt đi qua AC và AB và cắt AC tại D, AB tại E. Sao cho BE = DC.

Xét tam giác BEC và tam giác DCB có:

BE = DC ( chứng minh trên )

ˆB=ˆC( giả thiết )

Cạnh BC chung

=> Tam giác BEC = tam giác DCB ( c.g.c )

Vậy nếu ˆB=ˆCthì AB = AC ( đpcm )

 x³ -7x +6 
= x³ -x²+x²-x-6x+6 
= x²(x-1)+x(x-1)-6(x-1) 
= (x-1)(x² +x-6) 
= (x-1)(x²-2x+3x-6) 
=(x-1)(x-2)(x+3) 

dễ cực lun

3 k cho người đánh nhung

a³ + 2a² - 13a + 10

= a³ - a² + 3a² - 3a - 10a +10

= a²(a-1) + 3a(a-1) -10(a-1)

= (a-1)(a²+3a-10)

= (a-1)(a-2)(a+5)

\(m^{2} - 5 m + 6 = \left(\right. m - 2 \left.\right) \left(\right. m - 3 \left.\right)\)

m^3 - 6m^2 + 9m +2m-6
= m(m^2-6m+9) +2(m-3)
= m(m-3)+2(m-3)
=(m+2)(m-3)
ht nha

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(a+2b)³-b(2a+b)³

=(2a+b)³.(a-b)