2xy – x2 – y2 + 16 (Có 2xy ; x2 ; y2, ta liên tưởng đến HĐT (1) hoặc (2))

= 16 – (x2 – 2xy + y2)

= 42 – (x – y)2 (xuất hiện hằng đẳng thức (3))

= [4 – (x – y)][4 + (x - y)]

= (4 – x + y)(4 + 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) \(x^2-16+y^2+2xy\)

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

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

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

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

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

c) \(y^2+x^2+2xy-16=x^2+2xy+y^2-16\)

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

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

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

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

= [( x + y) - 10)] . ( x - y)

  \(x^2\) - y² - 10\(x\) - 10y 

= (\(x^2\) - y²) - 10.(\(x\) + y)

= (\(x-y\))(\(x\) + y) - 10.(\(x\) + y)

= (\(x\) + y).(\(x\) - y - 10)

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

Bạn thử xem lại đề câu d nhé.

a) Ta có: \(4x\left(2x-3y\right)-8y\left(3y-2x\right)\)

\(=4x\left(2x-3y\right)+8y\left(2x-3y\right)\)

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

b) Ta có: \(4x^2-4xy+y^2-9z^2\)

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

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

c) Ta có: \(x^2y+yz+xy^2+xz\)

\(=xy\left(x+y\right)+z\left(x+y\right)\)

\(=\left(x+y\right)\left(xy+z\right)\)

k cho tui nha

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