a\(^3\)-b\(^3\)+c\(^{^{ }3}\)+3abc

=a\(^3\)-b\(^3\)+c\(^3\)+3abc+3a\(^2\)b+3ab\(^2\)-3a\(^2\)b-3ab\(^2\)

=(a-b)\(^3\)+c\(^3\)+3ab(a-b+c)

=(a-b+c)[(a-b)\(^2\)-(a-b)c+c\(^2\)]+3ab(a-b+c)

=(a-b+c)(a\(^2\)+b\(^2\)+c\(^2\)-ab-ac+bc)

d,a\(^2\)-2a+2b-b\(^2\)

=(a-b)(a+b)-2(a-b)

=(a-b)(a+b-2)

c,5x\(^2\)+6xy+y\(^2\)

=5x(x+y)+y(x+y)

=(x+y)(5x+y)

b,16\(\)x\(^2\)-(x+y)\(^2\)

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

=((3x-y)(5x+y)

a,x\(^2\)-y\(^2\)-5x+5y

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

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

2,\(\dfrac{x-1}{x-2}\)+\(\dfrac{2x-3}{x-2}\)+\(\dfrac{x-4}{x-2}\)

=\(\dfrac{x-1+2x-3+x-4}{x-2}\)

=\(\dfrac{4x-8}{x-2}\)

=4

1,x\(^3\)+2x\(^2\)y+xy\(^2\)

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

=x(x+y)\(^2\)

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

=x(x-y)-4(x-y)

=(x-y)(x-4)

h,x\(^2\)-y\(^2\)+2x-2y

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

(x-y)(x+y+2)

bài 2

a,để phân thức được xác định thì x\(^3\)-x\(\ne\)0

=>x(x-1)(x+1)\(\ne\)0

=>x\(\ne\)0,x\(\ne\pm\)1

b,với x\(\ne\)0;x \(\ne\pm\)1 thì

P=\(\dfrac{x^3+2x^2+x}{x^3-x}\)

P=\(\dfrac{x\left(x+1\right)^2}{x\left(x-1\right)\left(x+1\right)}\)

P=\(\dfrac{x+1}{x-1}\)

d, 3x\(^2\)-3y\(^2\)-12x+12y

=3(x\(^2\)-y\(^2\)-6x+6y)

=3[(x-y)(x+y)-6(x-y)]

=3(x-y)(x+y-6)

e,x\(^2\)-3x-2

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

=(x-2)(x-1)

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

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

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

g,x\(^2\)-xy+7x-7y

=x(x-y)+7(x-y)

=(x-y)(x+7)

c,5x\(^3\)-x\(^2\)y-10x\(^2\)+10xy

=x(5x\(^2\)-xy-10x+10y)