a) \(xy+y-2x-2\)

\(=y\left(x+1\right)-2\left(x+1\right)\)

\(=\left(x+1\right)\left(y-2\right)\)

b) \(xy+1+x+y\)

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

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

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

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

Bài `1`

\(a,5x^2-10xy=5x\left(x-2y\right)\\ b,3x\left(x-y\right)-6\left(x-y\right)=\left(x-y\right)\left(3x-6\right)\\ =3\left(x-y\right)\left(x-2\right)\\ c,2x\left(x-y\right)-4y\left(y-x\right)=2x\left(x-y\right)+4y\left(x-y\right)\\ =\left(x-y\right)\left(2x+4y\right)=2\left(x-y\right)\left(x+2y\right)\\ d,9x^2-9y^2=\left(3x\right)^2-\left(3y\right)^2=\left(3x-3y\right)\left(3x+3y\right)\\ f,xy-xz-y+z=\left(xy-xz\right)-\left(y-z\right)\\ =x\left(y-z\right)-\left(y-z\right)=\left(y-z\right)\left(x-1\right)\)

Bài `3`

\(a,3x^2+8x=0\\ \Leftrightarrow x\left(3x+8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\3x+8=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\3x=-8\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{8}{3}\end{matrix}\right.\)

\(b,9x^2-25=0\\ \Leftrightarrow\left(3x\right)^2-5^2=0\\ \Leftrightarrow\left(3x-5\right)\left(3x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-5=0\\3x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=5\\3x=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)

\(c,x^3-16x=0\\ \Leftrightarrow x\left(x^2-16\right)=0\\ \Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)

\(d,x^3+x=0\\ \Leftrightarrow x\left(x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1\in\varnothing\\x=0\end{matrix}\right.\Rightarrow x=0\)

Lời giải:
a. $5x^2-10xy=5x(x-2y)$

b. $3x(x-y)-6(x-y)=(x-y)(3x-6)=3(x-y)(x-2)$
c. $2x(x-y)-4y(y-x)=2x(x-y)+4y(x-y)=(x-y)(2x+4y)=2(x-y)(x+2y)$

d. $9x^2-9y^2=9(x^2-y^2)=9(x-y)(x+y)$

e. $x^2-xy-x+y=(x^2-xy)-(x-y)=x(x-y)-(x-y)=(x-y)(x-1)$

f. $xy-xz-y+z=(xy-y)-(xz-z)=y(x-1)-z(x-1)=(x-1)(y-z)$

a, \(\left(2x+1\right)^2-2\left(2x+1\right)\left(x-3\right)+\left(x-3\right)^2\)

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

b, \(xy+xz+3y+3z=x\left(y+z\right)+3\left(y+z\right)=\left(x+3\right)\left(y+z\right)\)

c, \(xy-xz+y-z=x\left(y-z\right)+\left(y-z\right)=\left(x+1\right)\left(y-z\right)\)

d, \(x^2-xy-8x+8y=\left(x^2-xy\right)-\left(8x-8y\right)\)

\(=x\left(x-y\right)-8\left(x-y\right)=\left(x-8\right)\left(x-y\right)\)

e, \(x^2+2xy+y^2-xz-yz=\left(x^2+2xy+y^2\right)-\left(xz+yz\right)\)

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

f, \(25-4x^2-4xy-y^2=25-\left(4x^2+4xy+y^2\right)\)

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

1,

a, (2x + 1- x + 3)² = (x+4)²

b,\(x\left(y+z\right)+3\left(y+z\right)=\left(y+z\right)\left(x+3\right)\)

c, \(x\left(y-z\right)+\left(y-z\right)=\left(y-z\right)\left(x+1\right)\)

d,\(x\left(x-y\right)+8\left(y-x\right)\)=\(\left(x-y\right)\left(x-8\right)\)

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

f,\(25-\left(4x^2+4xy+y^2\right)=5^2-\left(2x+y\right)^2\)

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

Chúc các bn hc tốt

a) \(xy+y-2x-2\)

\(=y\left(x+1\right)-2\left(x+1\right)\)

\(=\left(x+1\right)\left(y-2\right)\)

b) \(xy+1+x+y\)

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

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

c) \(x^2+xy-x-y+xz-z\)

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

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

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

Ý a có rì đó sai sai nha bn 

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

x² - x - y² - y

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

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

***

9x² + y² - 16z² + 6xy

= (3x + y)² - (4z)²

= (3x + y - 4z)(3x + y + 4z)

***

a³ - a²x - ay + xy

= a²(a - x) - y(a - x)

= (a - x)(a² - y)

***

2x² - 8y² + 3x + 6y

= 2(x² - 4y²) + 3(x + 2y)

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

= (x + 2y)(2x - 4y + 3)

***

xy(x + y) + yz(y + z) + xz(x + z) + 2xyz

= xy(x + y + z) + yz(x + y + z) + xz(x + z)

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

= (x + z)(xy + y² + yz + xz)

= (x + z)[y(x + y) + z(x + y)]

= (x + z)(x + y)(y + z) 

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 

\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)