Lời giải:

a.

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

b.

$x^8-5x^4+4=(x^8-x^4)-(4x^4-4)$

$=x^4(x^4-1)-4(x^4-1)=(x^4-1)(x^4-4)$

$=(x^2-1)(x^2+1)(x^2-2)(x^2+2)$

$=(x-1)(x+1)(x^2+1)(x-\sqrt{2})(x+\sqrt{2})(x^2+2)$

c.

$x^3+3x^2+3x-7=(x^3+3x^2+3x+1)-8$

$=(x+1)^3-2^3=(x+1-2)[(x+1)^2+2(x+1)+4]$

$=(x-1)(x^2+4x+7)$

a) \(3x^2+xy-4y^2=3x^2-3xy+4xy-4y^2\)

\(=3x(x-y)+4y(x-y)=(3x+4y)(x-y)\)

b)\(x^8-5x^4+4=x^8-x^4-4x^4+4\)

\(=x^2(x^4-1)-4(x^4-1)=(x^2-4)(x^4-1)\)

\(=(x-2)(x+2)(x^2-1)(x^2+1)=(x-2)(x+2)(x-1)(x+1)(x^2+1)\)

c)\(x^3+3x^2+3x-7=x^3+3x^2+3x+1-8\)

\(\left(x+1\right)^3-\sqrt{2}^3=\left(x+1-\sqrt[]{2}\right)\left(\left(x+1\right)^2+2\sqrt{2}x+2\right)\)

a: \(3x^2+xy-4y^2\)

\(=3x^2+4xy-3xy-4y^2\)

\(=x\left(3x+4y\right)-y\left(3x+4y\right)\)

\(=\left(3x+4y\right)\left(x-y\right)\)

b: \(x^8-5x^4+4\)

\(=x^8-x^4-4x^4+4\)

\(=x^4\left(x^4-1\right)-4\left(x^4-1\right)\)

\(=\left(x^4-4\right)\left(x^4-1\right)\)

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

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

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

\(a,=-\left(x-1\right)^3\left[=\left(1-x\right)^3\right]\\ b,=\left(1-x\right)^3\)

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

b. \(=\left(1-x\right)^3\)

= 2x(3-4) + 4

= -2x + (-2).(-2)

= -2(x-2)

3x^2-8x+4

=3x^2-6x-2x+4

=3x(x-2)-2(x-2)

=(x-2)(3x-2)

l-kie cho mik nha bạn

x 8   +   x 4   +   1     =   x 8   +   2 x 4     +   1   –   x 4       =   ( x 8   +   2 x 4     +   1 )   –   x 4       =   [ ( x 4 ) 2   +   2 . x 4   . 1   +   12 ]   –   x 4     =   ( x 4   +   1 ) 2   –   ( x 2 ) 2     =   ( x 4   +   1   –   x 2 ) ( x 4   +   1   +   x 2 )     =   ( x 4   –   x 2   +   1 ) ( x 4   +   2 x 2   –   x 2   +   1 )     =   ( x 4   –   x 2   +   1 ) [ ( ( x 2 ) 2   +   2 . 1 . x 2   +   1 )   –   x 2 ] =   ( x 4   –   x 2   +   1 ) [ ( x 2   +   1 ) 2   –   x 2 ]     =   ( x 4   –   x 2   +   1 ) ( x 2   +   1   –   x ) ( x 2   +   1   +   x )     =   ( x 4   –   x 2   +   1 ) ( x 2   –   x   +   1 ) ( x 2   +   x   +   1 )

Đáp án cần chọn là: C

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

2x³ + 3x² - 2x

= x ( 2x² + 3x - 2 )

= x ( 2\(x^2\) + 4\(x-x-2\) )

= x [ ( 2\(x^2\) + 4x ) - ( x + 2 )]

= x  [  2x ( x + 2 ) - ( x + 2 )]

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

\(=12x^2-6x=6x\left(2x-1\right)\)

\(-9x^4+3x^2+2\\ =-9x^4+6x^2-3x^2+2\\ =-3x^2\left(3x^2-2\right)-\left(3x^2-2\right)\\ =-\left(3x^2-2\right)\left(3x^2+1\right)\)

\(-9x^4+3x^2+2\)

\(=-9x^4+6x^2-3x^2+2\)

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

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

Bạn phải vt thêm dấu mũ vào mới giải đc chứ!! Để thế kia ai mà giải đc

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

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

a) x³-3x²+3x-1-y³    =(x-1)³ - y³

= (x-1)³ - 3.(x-1)².y + 3.(x-1). y² - y