b)x(x^2-2x+1)

=x(x+1)^2

d)(3x^2-7x)-(3xy-7y)

=x(3x-7)-y(3x-7)

=(3x-7)(x-y)

\(a,=\left(x+1\right)\left(x+3\right)\\ b,=-5x^2+15x+x-3=\left(x-3\right)\left(1-5x\right)\\ c,=2x^2+2x+5x+5=\left(2x+5\right)\left(x+1\right)\\ d,=2x^2-2x+5x-5=\left(x-1\right)\left(2x+5\right)\\ e,=x^3+x^2-4x^2-4x+x+1=\left(x+1\right)\left(x^2-4x+1\right)\\ f,=x^2+x-5x-5=\left(x+1\right)\left(x-5\right)\)

Bài 1: 

a: \(=6x^3-10x^2+6x\)

b: \(=-2x^3-10x^2-6x\)

Bài 4: 

a: =>3x+10-2x=0

=>x=-10

c: =>3x²-3x²+6x=36

=>6x=36

hay x=6

Bài 1:

\(a,=6x^3-10x^2+6x\\ b,=-2x^3-10x^2-6x\)

Bài 4:

\(a,\Leftrightarrow3x+10-2x=0\Leftrightarrow x=-10\\ b,\Leftrightarrow x\left(2x^2+9x-5\right)-\left(2x^3+9x^2+x+4,5\right)=3,5\\ \Leftrightarrow2x^3+9x^2-5x-2x^3-9x^2-x-4,5=3,5\\ \Leftrightarrow-6x=8\Leftrightarrow x=-\dfrac{4}{3}\\ c,\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow x=6\)

Bài 1:

\(a,=7xy\left(2x-3y+4xy\right)\\ b,=x\left(x+y\right)-5\left(x+y\right)=\left(x-5\right)\left(x+y\right)\\ c,=\left(x-y\right)\left(10x+8\right)=2\left(5x+4\right)\left(x-y\right)\\ d,=\left(3x+1-x-1\right)\left(3x+1+x+1\right)\\ =2x\left(4x+2\right)=4x\left(2x+1\right)\\ e,=5\left[\left(x-y\right)^2-4z^2\right]=5\left(x-y-2z\right)\left(x-y+2z\right)\\ f,=x^2+8x-x-8=\left(x+8\right)\left(x-1\right)\\ g,\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\\ =\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\\ h,=x^2+3x+x+3=\left(x+3\right)\left(x+1\right)\)

a) 3a +3b -a²-ab

= 3.(a+b) -a.(a+b)=(3-a).(a+b)

b) x² +x +y²-y-2xy

=(x² - 2xy+y²) +(x-y)

=(x-y).(x-y+1)

c) -x² +7x -6

= -x² + x +6x-6

= x.(1-x) -6.(1-x) = (1-x).(x-6)

d) 5x³y -10x²y² +5xy³

= 5xy.(x² -2xy +y²) = 5xy.(x-y)²

e) 2x² +7x -15

= 2x² -3x +10x -15

=x.(2x-3) + 5.(2x-3)

=(2x-3).(x+5)

g) x² -2x +2y -xy

=x.(x-2)-y.(x-2)

=(x-y).(x-2)

h) bn go lai de ho mk dc k?

\(a,a^2-2a-4b^2-4b\)

\(=\left(a^2-4b^2\right)-\left(2a+4b\right)\)

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

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

\(b,x^3-2x^2+4x-8\)

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

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

\(c,x^3+36x-12x^2\)

\(=x^3-6x^2-6x^2+36x\)

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

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

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

\(d,5a^2+3\left(a+b\right)^2-5b^2\)

\(=\left(5a^2-5b^2\right)+3\left(a+b\right)^2\)

\(=5\left(a^2-b^2\right)+3\left(a+b\right)^2\)

\(=5\left(a-b\right)\left(a+b\right)+3\left(a+b\right)^2\)

\(=\left(a+b\right)\left[5\left(a-b\right)+3\left(a+b\right)\right]\)

\(=\left(a+b\right)\left(5a-5b+3a+3b\right)\)

\(=\left(a+b\right)\left(8a-2b\right)\)

\(=2\left(a+b\right)\left(4a-b\right)\)

\(e,x^3-3x^2+3x-1-y^3\)

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

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

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

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

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

#Urushi☕

\(c.\\ x^3+36x-12x^2\\ =x\left(x^2-12x+36\right)\\ =x.\left(x^2-2.x.6+6^2\right)\\ =x.\left(x-6\right)^2\\ ---\\ d.\\ 5a^2+3\left(a+b\right)^2-5b^2\\ =\left(5a^2-5b^2\right)+3\left(a+b\right)^2\\ =5.\left(a^2-b^2\right)+3.\left(a+b\right)\left(a+b\right)\\ =5\left(a+b\right)\left(a-b\right)+3\left(a+b\right)\left(a+b\right)\\ =\left(a+b\right)\left(5a-5b+3a+3b\right)\\ =\left(a+b\right)\left(8a-2b\right)\\ =2\left(a+b\right)\left(4a-b\right)\)

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

\(a.\\ a^2-2a-4b^2-4b\\ =\left(a^2-2a+1\right)-\left(4b^2-4b+1\right)\\ =\left(a-1\right)^2-\left(2b-1\right)^2\\ =\left[\left(a-1\right)+\left(2b-1\right)\right].\left[\left(a-1\right)-\left(2b-1\right)\right]\\ =\left(a+2b-2\right)\left(a-2b\right)\)

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

Bài 2 : phân tích các đa thức sau thành nhân tử

a, x3 - 2x2 + x

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

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

b, x2 - 2x - y2 + 1

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

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

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

vt mũ hộ mk đuy bạn :

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

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

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

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

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

\(b,x^2-2x-y^2+1\)

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

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

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

Bài 1 :

a) \(x\left(x^2+5\right)\)

\(=x^3+5x\)

\(2x^2-7x+5=\left(2x^2-2x\right)-\left(5x-5\right)=2x\left(x-1\right)-5\left(x-1\right)=\left(2x-5\right)\left(x-1\right)\)

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

a: \(2x^2-7x+5=\left(x-1\right)\left(2x-5\right)\)

b: \(3x^2+5x+2=\left(x+1\right)\left(3x+2\right)\)