c) x2 + 2xy + y2 – xz – yz = (x + y)2 – z(x + y) = (x + y)(x + y – z)

c) x2 + y2 + xz + yz + 2xy

= (x2 + 2xy + y2) + (xz + yz)

= (x + y)2 + z(x + y)

= (x + y)(x + y + z)

\(1,\\ a,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\\ b,=a^2\left(a-x\right)-y\left(a-x\right)=\left(a^2-y\right)\left(a-x\right)\\ c,=\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(x-y-z\right)\\ d,=x\left(x-2y\right)+t\left(x-2y\right)=\left(x+t\right)\left(x-2y\right)\\ 2,\\ \Rightarrow x^2-4x+4-x^2+9=6\\ \Rightarrow-4x=-7\Rightarrow x=\dfrac{7}{4}\\ 3,\\ a,x^2+2x+2=\left(x+1\right)^2+1\ge1>0\\ b,-x^2+4x-5=-\left(x-2\right)^2-1\le-1< 0\)

Bạn thử xem lại đề câu d nhé.

a) Ta có: \(4x\left(2x-3y\right)-8y\left(3y-2x\right)\)

\(=4x\left(2x-3y\right)+8y\left(2x-3y\right)\)

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

b) Ta có: \(4x^2-4xy+y^2-9z^2\)

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

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

c) Ta có: \(x^2y+yz+xy^2+xz\)

\(=xy\left(x+y\right)+z\left(x+y\right)\)

\(=\left(x+y\right)\left(xy+z\right)\)

a: \(x^2-y^2-x-y\)

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

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

f: \(x^3-5x^2-5x+1\)

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

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

\(a,=3xyz\left(x+2\right)\\ b,=5\left(x+2\right)-x\left(x+2\right)=\left(x+2\right)\left(5-x\right)\\ c,=\left(x+y\right)^2-z^2=\left(x+y-z\right)\left(x+y+z\right)\)

a) 3x²yz + 6xyz = 3xyz(x+2)
b) 5(x+2) - x² - 2x = 5(x+2) - x(x+2) = (5+x)(x+2)
c) x² + 2xy + y² - 2² = (x²+2xy+y²) - 2² = (x+y)² - 2² = (x+y+2)(x+y-2)

3x^2yz + 6xyz=3xyz(x+2)
5(x+2)-x^2-2x=5(x+2)-(x^2+2x)=5(x+2)-x(x+2)=(x+2)(5-x)
x^2+2xy+y^2-2^2=(x+y)^2 -2^2=(x+y+2)(x+y-2)

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).

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

a) 3xy² - 3x³ - 6xy + 3x 

=3x (y² - x² - 2y +1)

= 3x [ (y-1)² -x² ]

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

b) 3x² +11x+6

= 3 x² +9x +2x +6

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

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

c) -x³ - 4xy² + 4x²y +16x

= -x (x² + 4y² - 4xy -16 )

= -x [(x -2y)² - 4² ]

= -x(x-2y-4)(x-2y+4)