a) Cách 1.

Ta có 2xy + 3z + 6y + xz = (2xy + xz) + (3z + 6y)

= x(2 y + z)+3(z + 2 y) = (z + 2y)(x + 3).

Cách 2.

Ta có 2xy + 3z + 6y + xz = (2x1/ + 6y) + (3z + xz)

= 2y(x + 3) + z(3 + x) = (z + 2y)(x + 3).

b) Biến đổi được a 4   -   9 rt 3   +   a 2 -9a = (a- 9)a( a 2  +1).

c) Biến đổi được 3 x 2  + 5y - 3xy + (-5x) = (x - y)(3x - 5).

d) Biến đổi được  x 2  - (a + b)x + ab = (x- a)(x - b).

e) Ta có 4 x 2 - 4xy + y 2   –   9 t 2 =  ( 2 x   -   y ) 2   -   ( 3 t ) 2

= (2x - y - 3t )(2x - y + 31).

g) Ta có  x 3   -   3 x 2 y   +   3 xy 2   -   y 3   -   z 3

= ( x   -   y ) 3   -   z 3 = (x - y - z)( x 2   +   y 2   +   z 2  - 2xy + xz - yz).

h) Ta có x 2   -   y 2 + 8x + 6y+ 7 = ( x 2  +8x + 16) - ( y 2  - 6y+ 9)

= ( x   +   4 ) 2   - ( y - 3 ) 2  =(x-y + 7)(x + y + l).

a) \(=a\left(a^3-9a^2+a-9\right)=a\left[a^2\left(a-9\right)+\left(a-9\right)\right]\)

\(=a\left(a-9\right)\left(a^2+1\right)\)

b) \(=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)

c) \(=x\left(2y+z\right)+3\left(2y+z\right)=\left(2y+z\right)\left(x+3\right)\)

d) \(=x^2-ax-bx+ab=x\left(x-a\right)-b\left(x-a\right)\)

\(=\left(x-a\right)\left(x-b\right)\)

a) = a(a³-9a²+a-9)

b) =3x²+5y-3xy-5x

= (3x²-5x)+(5y-3xy)

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

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

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

c)2xy +3z+6y+xz

=(2xy+6y)+(3z+xz)

=2y(x+3)+z(3+x)

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

a) \(a^4-9a^3+a^2-9a=a\left(a^3-9a^2+a-9\right)=a\left[\left(a^3-9a^2\right)+\left(a-9\right)\right]=a\left(a^2+1\right)\left(a-9\right)\)

\(a.2x^3+6x=2x\left(x^2+3\right)\)

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

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

\(b.5x\left(x-2\right)-3x^2\left(x-2\right)\)

\(=\left(x-2\right)\left(5x-3x^2\right)\)

\(c.3x\left(x-5y\right)-2y\left(5y-x\right)\)

\(=3x\left(x-5y\right)+2\left(x-5y\right)\)

\(=\left(x-5y\right)\left(3x+2\right)\)

\(d.y^2\left(x^2+y\right)-x^3-xy\)

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

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

e. Cái bài này ghi lại đàng hoàng xíu nha t k hỉu

\(f.3x^2\left(y^2-2x\right)-15x\left(2x-y^2\right)\)

\(=3x^2\left(y^2-2x\right)+15x\left(y^2-2x\right)\)

\(=\left(y^2-2x\right)\left(3x^2+15x\right)\)

Cái này y hệt cái đề mik thi:)

trời dài thế làm lâu phết đó nha hừm làm theo đúng công thức là được :)

b) 24x^2+6x^2y−2x−12yx−3y^2x+y 

tôi làm theo cách tìm tích số

nếu thấy đúng thì tick cho tôi nha

a.

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

b.

\(8-27x^3=\left(2\right)^3-\left(3x\right)^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)

c.

\(27+27x+9x^2+x^3=x^3+3.x^2.3+3.3^2.x+3^3\)

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

d.

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

e.

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

\(=\left(x-y\right)\left(x+y-5\right)\)

f.

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

g. 10x(x-y)-6y(y-x)

=10x(x-y)+6y(x-y)

=(x-y)(10x+6y)

h.x²-4x-5

=(x-5)(x+1)

i.x⁴-y⁴ = (x²-y²)(x²+y²)

B2.

a.5(x-2)=x-2

⇔5(x-2)-(x-2)=0

⇔4(x-2)=0

⇔x=2

b.3(x-5)=5-x

⇔3(x-5)+(x-5)=0

⇔4(x-5)=0

⇔x=5

c.(x+2)²-(x+2)(x-2)=0

⇔(x+2)[(x+2)-(x-2)]=0

⇔4(x+2)=0

⇔x=-2

a) \(14x^2y-21xy^2+28x^2y^2\)

\(=7xy\left(2x-3y+4xy\right)\)

b) \(3x^2-5x-3xy+5y\)

\(=\left(3x^2-3xy\right)-\left(5x-5y\right)\)

\(=3x\left(x-y\right)-5\left(x-y\right)\)

\(=\left(x-y\right)\left(3x-5\right)\)

c) \(5a^3-20a\)

\(=5a\left(a^2-4\right)\)

\(=5a\left(a-2\right)\left(a+2\right)\)

d) \(2x+2y+x^2+2xy+y^2\)

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

= \(=\left(x+y\right)\left(2+x+y\right)\)

\(a,10x^2y-20xy^2=10xy\left(x-2y\right)\\ b,x^2-y^2+10y-25=x^2-\left(y^2-10y+25\right)=x^2-\left(y-5\right)^2=\left(x-y+5\right)\left(x+y-5\right)\\ c,x^2-y^2+3x-3y=\left(x-y\right)\left(x+y\right)+3\left(x-y\right)=\left(x-y\right)\left(x+y+3\right)\\ d,x^3+3x^2-16x-48=\left(x^3+3x^2\right)-\left(16x+48\right)=x^2\left(x+3\right)-16\left(x+3\right)=\left(x+3\right)\left(x^2-16\right)=\left(x+3\right)\left(x+4\right)\left(x-4\right)\)

\(e,9x^3+6x^2+x=x\left(9x^2+6x+1\right)=x\left(3x+1\right)^2\\ f,x^4+5x^3+15x-9=\left(x^4+5x^3-3x^2\right)+\left(3x^2+15x-9\right)=x^2\left(x^2+5x-3\right)+3\left(x^2+5x-3\right)=\left(x^2+3\right)\left(x^2+5x-3\right)\)