2x – 2y – x2 + 2xy – y2

(Có x2 ; 2xy ; y2 ta liên tưởng đến HĐT (1) hoặc (2))

= (2x – 2y) – (x2 – 2xy + y2)

= 2(x – y) – (x – y)2

(Có x – y là nhân tử chung)

= (x – y)[2 – (x – y)]

= (x – y)(2 – x + y)

a) \(x^2-y^2-3x+3y\)

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

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

b) \(2x+2y-x^2+y^2\)

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

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

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

c) \(x^2-16+y^2+2xy\)

\(=x^2+y^2+2xy-16\)

\(=\left(x+y\right)^2-16\)

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

a) \(x^2-y^2-3x+3y\)

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

b) \(2x+2y-x^2+y^2\)

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

c) \(x^2-16+y^2+2xy\)

\(=\left(x+y\right)\left(x-y\right)+2xy-16\)

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

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

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

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

2x - 2y - x² + 2xy - y²

= (2x - 2y) - (x² - 2xy + y²)

= 2(x - y) - (x - y)²

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

a) x²-2x-y²+2y

=(x²-y²)-(2x-2y)

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

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

d) x²-25+y²+2xy

=(x²+y²+2xy)-5²

=(x+y)²-5²

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

`x^2+2x+1-y^2+2y-1`

`=(x^2+2x+1)-(y^2-2y+1)`

`=(x+1)^2-(y-1)^2`

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

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

Ta có: \(x^2+2x+1-y^2+2y-1\)

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

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

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

a) \(14x^2y-21xy^2+28x^2y^2\)

\(=7xy\left(2x-3y+4xy\right)\)

b) \(3x^2-5x-3xy+5y\)

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

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

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

c) \(5a^3-20a\)

\(=5a\left(a^2-4\right)\)

\(=5a\left(a-2\right)\left(a+2\right)\)

d) \(2x+2y+x^2+2xy+y^2\)

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

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

\(a,=x\left(x-3\right)\\ b,=\left(x-y\right)^2-4=\left(x-y-2\right)\left(x-y+2\right)\\ c,=\left(3x+7\right)\left(x-2\right)+2\left(x-2\right)=\left(x-2\right)\left(3x+9\right)=3\left(x+3\right)\left(x-2\right)\)

c) \(5x^2+3y+15x+xy=5x\left(x+3\right)+y\left(x+3\right)=\left(x+3\right)\left(5x+y\right)\)

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

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

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

c: \(5x^2+15x+3y+xy\)

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

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

d: \(x^2+6x+9-y^2\)

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

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

e: \(x^2+2x+1-y^2\)

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

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

f: \(x^2-2xy+y^2-9\)

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

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

Lời giải:
$\frac{x}{y}$ không phải đơn thức bạn nhé.

a. $x^2-2x+1=(x-1)^2$

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

c. $5x^2-10xy=5x(x-2y)$

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

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