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

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

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

b) \(\left(x-5\right)^2-9\left(x+y\right)^2\)

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

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

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

\(=\left(4x+3y-5\right)\left(-2x-3y-5\right)\)

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

e: \(x^2-10x+24=\left(x-4\right)\left(x-6\right)\)

g) \(6x^2+7x-5\)

=\(6x^2+10x-3x-5\)

=\(\left(6x^2+10x\right)-\left(3x+5\right)\)

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

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

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).

\(b,=x^2-xy=x\left(x-y\right)\\ c,=y\left(x+1\right)+z\left(x+1\right)=\left(y+z\right)\left(x+1\right)\\ d,=x^2+ax+bx+ab\\ =x\left(x+a\right)+b\left(x+a\right)=\left(x+b\right)\left(x+a\right)\)

A = xy + y - 2x - 2

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

= ( x + 1 )( y - 2 )

B = x² - 3x + xy - 3y

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

= ( x - 3 )( x + y )

C = 3x² - 3xy - 5x + 5y

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

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

D = xy + 1 + x + y

= y( x + 1 ) + ( x + 1 )

= ( x + 1 )( y + 1 )

E = ax - bx + ab - x²

= ( ax - x² ) + ( ab - bx )

= x( a - x ) + b( a - x )

= ( a - x )( x + b )

F = x² + ab + ax + bx

= ( ax + x² ) + ( ab + bx )

= x( a + x ) + b( a + x )

= ( a + x )( x + b )

G = a³ - a²x - ay + xy

= a²( a - x ) - y( a - x )

= ( a - x )( a² - y )

Bonus : = ( a - x )[ a² - ( √y )² ]

             = ( a - x )( a - √y )( a + √y )

H = 2xy + 3z + 6y + xz

= ( 6y + 2xy ) + ( 3z + xz )

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

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

A = xy + y - 2x - 2 = y(x + 1) - 2(x + 1) = (y - 2)(x + !1

B = x² - 3x + xy - 3y = x(x - 3) + y(x - 3) = (x + y)(x - 3)

C = 3x² - 3xy - 5x + 5y = 3x(x - y) - 5(x - y) = (3x - 5)(x - y)

D = xy + 1 + x + y = xy + x + y + 1 = x(y + 1) + (y + 1) = (x + 1)(y + 1)

E = ax - bx + ab - x² = ax - x² + ab - bx = a(a - x) - b(a - x) = (a - b)(a - x)

F = x² + ab + ax + bx = ab + ax + bx + x² = a(b + x) + x(b + x) = (a + x)(b + x)

G = a³ - a²x - ay + xy = a²(a - x) - y(a - x) = (a² - y)(a - x)

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

🍀Trần Nhật Quỳnh🍀 y không dương nên không thể cho vào căn nhé 

Bài 4: 

Ta có: \(\left(x^3-x^2\right)-4x^2+8x-4=0\)

\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

từng câu 1 thôi:v

a) x2-xy+5y-25
 = x(2-y)+ 5(y-2)
 = x(2-y)-5(2-y)
 = (x-5)(2-y)

h: \(=\left(a-1-b\right)\left(a-1+b\right)\)

a Đề sai: )

b

\(a^3-a^2x-ay+xy\\ =a^2\left(a-x\right)-y\left(a-x\right)\\ =\left(a-x\right)\left(a^2-y\right)\)

c

\(4x^2-y^2+4x+1\\ =\left(2x\right)^2+2.2x.1+1-y^2\\ =\left(2x+1\right)^2-y^2\\ =\left(2x+1-y\right)\left(2x+1+y\right)\)

d

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

e

\(5x^2-10xy+5y^2-5z^2\\ =5\left(x^2-2xy+y^2-z^2\right)\\ =5\left[\left(x-y\right)^2-z^2\right]\\ =5\left(x-y-z\right)\left(x-y+z\right)\)

c: =(2x+1)^2-y^2

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

d: =x^2(x^2+2x+1)

=x^2(x+1)^2

e: =5(x^2-2xy+y^2-z^2)

=5[(x-y)^2-z^2]

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