a: Ta có: \(10x^4-27x^3y-110x^2y^2-27xy^3+10y^4\)

\(=10x^4+20x^2y^2+10y^4-27xy\left(x^2+y^2\right)-130x^2y^2\)

\(=10\left(x^2+y^2\right)^2-27xy\left(x^2+y^2\right)-130x^2y^2\)

\(=10\left(x^2+y^2\right)^2-52xy\left(x^2+y^2\right)+25xy\left(x^2+y^2\right)-130x^2y^2\)

\(=2\left(x^2+y^2\right)\left(5x^2+5y^2-26xy\right)+5xy\left(5x^2+5y^2-26xy\right)\)

\(=\left(5x^2-26xy+5y^2\right)\left(2x^2+5xy+2y^2\right)\)

\(=\left(5x^2-25xy-xy+5y^2\right)\left(2x^2+4xy+xy+2y^2\right)\)

\(=\left\lbrack5x\left(x-5y\right)-y\left(x-5y\right)\right\rbrack\left\lbrack2x\left(x+2y\right)+y\left(x+2y\right)\right\rbrack\)

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

b: \(x^5-4x^4+3x^3+3x^2-4x+1\)

\(=x^5+x^4-5x^4-5x^3+8x^3+8x^2-5x^2-5x+x+1\)

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

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

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

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

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

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!!!

4x^4+y^4= 4x^4 +4x^2y^2+y^4-4x^2y^2= ( 2x^2 + y^2 ) ^2 - ( 2xy ) ^2

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

haha lớp trưởng lớp tôi mà cux không làm đc câu này cơ đấy.....

helpppppp

a, x^2-4x+3

=x^2-x-3x+3

=x(x-1)-3(x-1)

=(x-3)(x-1)

\(\left(x^2+x\right)^2+4x^2+4x-12\)

\(=x^4+2x^3+5x^2+4x-12\)

\(=x^4-x^3+3x^3-3x^2+8x^2-8x+12x-12\)

\(=x^3.\left(x-1\right)+3x^2.\left(x-1\right)+8x.\left(x-1\right)+12.\left(x-1\right)\)

\(=\left(x-1\right).\left(x^3+3x^2+8x+12\right)=\left(x-1\right).\left(x+2\right).\left(x^2+x+6\right)\)

p/s: sai sót bỏ qua

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) 

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)

dễ mak

nếu dễ thì trả lời hộ đi