cái này dễ mà

= (2x)^3-3(2x)^2*1+2*3x*1^2-1^3

= (2x-1)^3

Câu a:

\(x^2-y^2-x+3y-2=\left(x^2-2.x.\frac{1}{2}+\frac{1}{4}\right)-\left(y^2-2.y.\frac{3}{2}+\frac{9}{4}\right)\)

\(< =>\left(x-\frac{1}{2}\right)^2-\left(y-\frac{3}{2}\right)^2\)

\(< =>\left(x-\frac{1}{2}+y-\frac{3}{2}\right)\left(x-\frac{1}{2}-y+\frac{3}{2}\right)=\left(x+y-2\right)\left(x-y+1\right)\)

a: Sửa đề: x^3-x^2+5x-5

=x^2(x-1)+5(x-1)

=(x-1)(x^2+5)

b: x^3+4x^2+x-6

=x^3-x^2+5x^2-5x+6x-6

=(x-1)(x^2+5x+6)

=(x-1)(x+2)(x+3)

c: \(=\left(x+2\right)^3+y^3\)

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

a) x^3 +5x^2+6x

= x^3+2x^2+3x^2+6x

=x*(x+3)*(x+2)

b) x^2-6x+8

= x^2-2x-4x+8

=(x-2)*(x-4)

c)2x^2+98+28x-8y^2

=2(x^2+14x+49-4y^2)

=2*[(x+7)^2-4y^2]

2*(x-7-2y)*(x-7+2y)

a) x² + 2x² + 3x² + 6x

= x( x+2 ) + 3( x+2 )

=(x+3)(x+2)

b) x² - 2x - 4x + 8

=x(x-2)-4(x-2)

=(x-4)(x-2)

c) 

a) x³+5x²+6x

= x³+2x²+3x²+6x

= (x³+2x²)+(3x²+6x)

= x²(x+2)+3x(x+2)

= (x²+3x)(x+2)

= x(x+3)(x+2)

b) x²-6x+8

= x²-2x-4x+8

= (x²-2x)-(4x-8)

= x(x-2)-4(x-2)

= (x-4)(x-2)

c) 2x²+98+28x+8y²

 = 2(x²+49+14x+4y²)

 = 2[(x²+14x+49)+4y²]

 = 2[(x+7)^²+4y²]

 = 2(x+7+2y)(x+7-2y)

đề có sai ko bn?

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

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

K sai dau
giao an truong Tran dai nghia do

tu de bai suy ra: 3x(x^3+4x+4)=3x(x+2)^2

3x^3+12x^2+12x=3x(x^2+4x+4)=3x(x+4)^2

zễ mà

1,2 áp dụng hằng đẳng thức   

3)(2x+y)³

dùng hằng đẳng thức để phân tích:

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

\(=\left(a+b+a-b\right)\left(a^2+2ab+b^2+b^2-a^2+a^2-2ab+b^2\right)\)

\(=2a\left(a^2+3b^2\right)\)

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

\(=\left(a+b+a-b\right)\left(a^2+2ab+b^2+a^2-b^2+a^2-2ab+b^2\right)\)

\(=2a\left(3a^2+b^2\right)\)

3)\(8x^3+12x^2y+6xy^2+y^3=\left(2x\right)^3+3.\left(2x\right)^2.y+3.2x.y^2+y^3=\left(2x+y\right)^3\)