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

c.3x²+5y-3xy-5x=(3x²--3xy)-(5x-5y)=3x(x-y)-5(x-y)=(3x-5)(x-y)

a.3x²-6x+9x²

= 3x(x-2+3x)

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

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

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

c.3x²+5y-3xy-5x

=(3x²-3xy)+(5y-5x)

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

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

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

d.3y²-3z²+3x²+6xy

=3(y²-z²+x²+2xy)

= 3(x²+2xy+y²)-z²

=3(x+y)²-z²

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

e.16x³+54y³

= 2(8x³+27y³)

=2(8x+27y)(8x²-8x.27y+27y²)

=2(8x+27y)(8x²-216xy+27y²)

f.x²-25-2xy+y²

= (x²-2xy+y²)-25

=(x-y)²-25

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

tớ ko làm đúng chính xác đâu nha

^-^

\(10x\left(x-y\right)-6y\left(y-x\right)\)

\(=10x\left(x-y\right)+6x\left(x-y\right)\)

\(=\left(10x+6x\right)\left(x-y\right)\)

\(c,3x^2+5y-3xy-5x\)

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

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

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

\(e,27+27x+9x^2=3\left(9+9x+x^2\right)\)

\(f,8x^3-12x^2y+6xy^2-y^3\)

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

\(g,x^3+8y^3=x^3+\left(2y\right)^3\)

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

\(i,x^2-25-2xy+y^2\)

\(\left(x^2-2xy+y^2\right)-25=\left(x-y\right)^2-5^2\)

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

a) \(3x^2-6x+9x^2\)

\(=12x^2-6x\)

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

b) hỏi chấm đề ?

c) \(3x^2+5y-3xy-5x\)

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

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

d) \(3y^2-3z^2+3x^2-6xy\)

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

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

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

e) \(16x^3+54y^3\)

\(=2\left(8x^3+27y^3\right)\)

\(=2\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)

f) \(x^2-25-2xy+y^2\)

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

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

g) \(x^5-3x^4-3x^3-x^2\)

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

a.\(xz+yz-5\left(x+y\right)\)

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

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

b.\(3x^2-3xy-5x+5y\)

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

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

c.\(x^2+6x-y^2-3z^2\)???Sai đề bài ...?

d.\(3x^2+6xy+3y^2-3z^2\)

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

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

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

Trả lời:

a, xz + yz - 5 ( x + y )

= ( xz + yz ) - 5 ( x + y )

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

= ( x + y ) ( z - 5 )

b, 3x² - 3xy - 5x + 5y

= ( 3x² - 3xy ) - ( 5x - 5y )

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

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

c, x² + 6x - y² - 3z²

= - ( 3x² - x² + y² - 6x )

d, 3x² + 6xy + 3y² - 3z²

= 3 ( x² + 2xy + y² - x² )

= 3 [ ( x² + 2xy + y² ) - z² ]

= 3 [ ( x + y )² - z² ]

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

Ai trả lời mình đi cho đỡ quê 😞💔

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

c: \(=\left(x-y\right)^2\)

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

\(3x^2+6xy+3y^2-3z^2=3\left(x^2+2xy+y^2-z^2\right)=3.\left[\left(x+y\right)^2-z^2\right]=3.\left(x+y-z\right)\left(x+y+z\right)\)

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

Lời giải:

a. $A=9x^2+15x+6xy+y^2+5y=(9x^2+6xy+y^2)+(15x+5y)$
$=(3x+y)^2+5(3x+y)=0^2+5.0=0$

b. $25x^2-y^4-5x+y^2=(25x^2-y^4)-(5x-y^2)=(5x-y^2)(5x+y^2)-(5x-y^2)$

$=(5x-y^2)(5x+y^2-1)$