a. xy+x+8y+8

= x(y+1)+8(y+1)

= (y+1)(x+8)

b. x²-x-2/3x+2/3

= x(x-1)-2/3(x-1)

= (x-1)(x-2/3)

c. x²-1

= x²-1²

= (x+1)(x-1)

a) xy + x + 8y + 8

= x.(y + 1) + 8.(y + 1)

= (y + 1).(x + 8)

b) \(x^2-x-\frac{2}{3}.x+\frac{2}{3}\)

\(=x.\left(x-1\right)-\frac{2}{3}.\left(x-1\right)\)

\(=\left(x-1\right).\left(x-\frac{2}{3}\right)\)

c) x² - 1

= x² + x - x - 1

= x.(x + 1) - (x + 1)

= (x + 1).(x - 1)

a) (xy+x) +(8y+8)=x(y+1)+8(y+1)=(x+8)(y+1)

b) (x²-x) -(2/3x-2/3)=x(x-1)+2/3(x-1)=(x+2/3)(x-1)

c) x²-1= (x-1)(x+1)

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

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

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

\(\left(x-y\right)\times\left(x-y\right)\)

ta có:

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

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

\(=\left[x.\left(x-y\right)\right]-\left[y.\left(x-y\right)\right]\)

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

\(a\left(x-1\right)+b\left(1-x\right)=a\left(x-1\right)-b\left(x-1\right)=\left(x-1\right)\left(a-b\right)\)

a) ( x + 1) + b ( 1 - x )

= ( x + 1 ) - b( x-   1 )

= ( 1 - b )( x- 1 )

**** 

a(x-1) + b(1-x)

= a(x-1) - b(x-1)

= (a-b)(x-1)

Nhấn đúng cho mk nha!!!!!!!!!

\(x^2-1=\left(x+1\right)\left(x-1\right)\)(Nhớ k cho mình với nhé!)

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

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

i) \(10x^2\left(a-2b\right)^2-\left(x^2+2\right)\left(2b-a\right)^2=10x^2\left(a-2b\right)^2-\left(x^2+2\right)\left(a-2b\right)^2=\left(a-2b\right)^2\left(10x^2-x^2-2\right)=\left(a-2b\right)^2\left(9x^2-2\right)\)

x² - 1 = x² - x + x - 1 = x . (x - 1) + (x - 1) = (x - 1) . (x + 1)

1.

\(a,\left(-xy\right)\left(-2x^2y+3xy-7x\right)\)

\(=2x^3y^2-3x^2y^2+7x^2y\)

\(b,\left(\dfrac{1}{6}x^2y^2\right)\left(-0,3x^2y-0,4xy+1\right)\)

\(=-\dfrac{1}{20}x^4y^3-\dfrac{1}{15}x^3y^3+\dfrac{1}{6}x^2y^2\)

\(c,\left(x+y\right)\left(x^2+2xy+y^2\right)\)

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

\(=x^3+3x^2y+3xy^2+y^3\)

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

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

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

2.

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

\(=x^3-y^3\)

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

\(=x^3+y^3\)

\(c,\left(4x-1\right)\left(6y+1\right)-3x\left(8y+\dfrac{4}{3}\right)\)

\(=24xy+4x-6y-1-24xy-4x\)

\(=\left(24xy-24xy\right)+\left(4x-4x\right)-6y-1\)

\(=-6y-1\)

#Toru

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