ai jup mjk s

bn xem lại đề xem hình như 3zy phải là 3z chứ 

3zy bạn ơi

Bạn ơi sai đề ko vậy????

ko bạn ơi thầy in đề vậy đó

Đề bài sai rồi bạn nhé ~

Tham khảo : https://olm.vn/hoi-dap/detail/92085650467.html

a/\(\left(x^2-3xy+\frac{9}{4}y^2+2x-\frac{9}{2}y+1\right)+\left(-\frac{9}{4}y^2+\frac{9}{2}y-1\right)=\left(x-\frac{3}{2}y+1\right)^2-\left(\frac{3}{2}y-1\right)^2=\left(x-\frac{3}{2}y+1-\frac{3}{2}y+1\right)\left(x-\frac{3}{2}y+1+\frac{3}{2}-1\right)=\left(x-3y+2\right)x\)b/\(=\left(x-y\right)^2-4=\left(x-y-2\right)\left(x-y+2\right)\)

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

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

\(N=x^2-3x-2xy+y^2+3y=\left(x-y\right)\left(x-y-3\right)\)\(K=2xy+3z+6y+xz=\left(x+3\right)\left(2y+z\right)\)

M= x²-5x+xy-5y= x(x-5)+y(x-5)=(x-5)(x+y)

N= x²-3x-2xy+y²+3y=(x-y)²-3(x-y)=(x-y)(x-y-3)

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

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) \(6x-6y=6\left(x-y\right)\)

b)\(2xy+3x+6y+xz\)

\(=\left(2xy+xz\right)+\left(6y+3z\right)\)

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

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

c)\(x^2+6x+9-y^2\)

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

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

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

d) \(9x-x^3\)

\(=x\left(9-x^2\right)\)

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

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

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

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

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

a, 6x - 6y = 6( x-y )

b, 2xy + 3z + 6y + xz

  = ( 2xy + 6y ) + ( 3z + xz )

  = 2y( x + 3 ) + z ( 3 + x )

  = 2y( 3 + x ) + z ( 3 + x )

  = ( 3 + x ) ( 2y + z )

c, x² + 6x + 9 - y² = ( x² + 6x + 9 ) - y²

                             = ( x + 3 )² - y²

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

d , 9x - x³ = x ( 9 - x² )

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

e, x² - xy + x - y =( x ² - xy ) + ( x - y )

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

                          = ( x - y ) ( x + 1 )

\(a,6x-6y=6\left(x-y\right)\)

\(b.2xy+3z+6y+xz\)

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

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

\(c,x^2+6x+9-y^2\)

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

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

\(d,9x-x^3\)

\(x\left(9-x^2\right)\)

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

\(e,x^2-xy+x-y\)

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

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

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)

a) 2xy + 3zy + 6y + xz

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

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

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

b) x² - 10x + 25

= x² - 2.x.5 + 5²

= (x - 5)²

c) x² + 6x + 9 - y²

= x² + 2.x.3 + 3² - y²

= (x - 3)² - y²

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

d) x³ - 4x² - xy² + 4x

= x(x² - 4x + 4 - y²)

= x[(x - 2)² - y²]

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