\(a,=\left(3-x+1\right)\left(9+3x-3+x^2-2x+1\right)\\ =\left(4-x\right)\left(x^2+x+7\right)\\ b,=4x^2-4xy-13xy+13y^2\\ =4x\left(x-y\right)-13y\left(x-y\right)\\ =\left(4x-13y\right)\left(x-y\right)\\ c,=4\left(x^2-xy-2y^2\right)\\ =4\left(x^2+xy-2xy-2y^2\right)\\ =4\left(x+y\right)\left(x-2y\right)\\ d,=x^3+4x^2+5x^2+20x+6x+24\\ =\left(x+4\right)\left(x^2+5x+6\right)\\ =\left(x+4\right)\left(x^2+2x+3x+6\right)\\ =\left(x+4\right)\left(x+2\right)\left(x+3\right)\\ f,=x\left(x+4y\right)-3\left(x+4y\right)=\left(x-3\right)\left(x+4y\right)\\ g,=4x^3+4x^2-29x^2-29x-24x-24\\ =\left(x+1\right)\left(4x^2-29x-24\right)\\ =\left(x+1\right)\left(4x^2-32x+3x-24\right)\\ =\left(x+1\right)\left(x-8\right)\left(4x+3\right)\)

\(a,27-\left(x-1\right)^3=\left(3-x+1\right)\left[9+3\left(x-1\right)+\left(x+1\right)^2\right]=\left(4-x\right)\left(9+3x-3+x^2+2x+1\right)=\left(4-x\right)\left(x^2+5x+7\right)\)

\(b,4x^2-17xy+13y^2=\left(4x^2-4xy\right)-\left(13xy-13y^2\right)=4x\left(x-y\right)-13y\left(x-y\right)=\left(x-y\right)\left(4x-13y\right)\)

\(c,4x^2-4xy-8y^2=4\left(x^2-xy-2y^2\right)\)

\(d,x^3+9x^2+26x+24=\left(x^3+2x^2\right)+\left(7x^2+14x\right)+\left(12x+24\right)=\left(x+2\right)\left(x^2+7x+12\right)=\left(x+2\right)\left[\left(x^2+3x\right)+\left(4x+12\right)\right]=\left(x+2\right)\left(x+3\right)\left(x+4\right)\)

\(f,4xy+x^2-3x-12y=x\left(4y+x\right)-3\left(x+4y\right)=\left(x+4y\right)\left(x-3\right)\)

\(g,4x^3-25x^2-53x-24=\left(4x^3-32x^2\right)+\left(7x^2-56x\right)+\left(3x-24\right)=\left(4x^2+7x+3\right)\left(x-8\right)=\left[\left(4x^2+4x\right)+\left(3x+3\right)\right]=\left(4x+3\right)\left(x+1\right)\left(x-8\right)\)

a) x²+x-2
= x²-x+2x-2
= x(x-1)+2(x-1)
= (x+2)(x-1)
b) 2x²+5x+3
= 2x²+2x+3x+3
= 2x(x+1)+3(x+1)
= (2x+3)(x+1)
c) 3x²+5x-2
= 3x²+6x-1x-2
= 3x(x+2)-1(x+2)
= (3x-1)(x+2)

a: \(x^4+3x^3+x^2+3x\)

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

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

c: \(x^2-xy-x+y\)

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

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

Ta có: \(\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4\)

\(=\left(12x^2+8x+3x+2\right)\left(12x^2+12x-x-1\right)-4\)

\(=\left(12x^2+11x+2\right)\left(12x^2+11x-1\right)-4\)

\(=\left(12x^2+11x\right)^2+\left(12x^2+11x\right)-6\)

\(=\left(12x^2+11x+3\right)\left(12x^2+11x-2\right)\)

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

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

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

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

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

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

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

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

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

a) $x^3-3x^2y+4x-12y$

$=(x^3-3x^2y)+(4x-12y)$

$=x^2(x-3y)+4(x-3y)$

$=(x-3y)(x^2+4)$

b) $4x^2-y^2+4y-4$

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

$=(2x)^2-(y^2-2\cdot y\cdot2+2^2)$

$=(2x)^2-(y-2)^2$

$=[2x-(y-2)][2x+(y-2)]$

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

c) $9x^2-6x-y^2+2y$

$=(9x^2-y^2)-(6x-2y)$

$=[(3x)^2-y^2]-2(3x-y)$

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

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

$\text{#}Toru$

bạn ấn ở chỗ x² cho rõ hơn nhé

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

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

\(=\left(x^2+4x+8\right)\left(x^2+6x+8\right)+x\left(x^2+6x+8\right)\)

\(=\left(x^2+4x+8+x\right)\left(x^2+6x+8\right)=\left(x^2+5x+8\right)\left(x^2+6x+8\right)\)

Ta có: \(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2\)

\(=\left(x^2+5x+8\right)\left(x^2+6x+8\right)\)

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

\(x^2-4x+3\)

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

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

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

x²-4x+3

=x.x-x-3x-3

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

=(x-1)(x-3)

x² - 4x + 3

= x² - x - 3x + 3 ( dùng phương pháp tách nhân tử )

= ( x² - x ) - ( 3x - 3 ) ( vì đằng trc có đấu " - " nên ta đổi đáu trong ngoặc )

= x( x - 1 ) - 3(x - 1 ) ( đặt nhân tử chung )

= ( x - 1 ) ( x - 3 )

hok tốt !