`(ax+by)^2-(ay+bx)^2`

`=(ax+by+ay+bx)(ax+by-ay-bx)`

`=[a(x+y)+b(x+y)][a(x-y)-b(x-y)]`

`=(x+y)(a+b)(x-y)(a-b)`

\(\left(ax+by\right)^2-\left(ay+bx\right)^2\)

\(=\left(ax-ay-bx+by\right)\left(ax+ay+bx+by\right)\)

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

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

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

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

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

\(\left(x+y\right)^2+3\left(x+y\right)-10=\left[\left(x+y\right)^2+2\left(x+y\right).\dfrac{3}{2}+\dfrac{9}{4}\right]-\dfrac{49}{4}\)

\(=\left(x+y+\dfrac{3}{2}\right)^2-\dfrac{49}{4}=\left(x+y+\dfrac{3}{2}-\dfrac{7}{2}\right)\left(x+y+\dfrac{3}{2}+\dfrac{7}{2}\right)=\left(x+y-2\right)\left(x+y+5\right)\)

\(\left(x+y\right)^2+3\left(x+y\right)-10\)

\(=\left(x+y\right)^2+5\left(x+y\right)-2\left(x+y\right)-10\)

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

\(=x\left[x^2\left(x-y\right)^2-36y^2\right]\\ =x\left[x\left(x-y\right)-6y\right]\left[x\left(x-y\right)+6y\right]\\ =x\left(x^2-xy-6y\right)\left(x^2-xy+6y\right)\)

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

Ta có:

 (x + y + z)² + (x + y – z)² – 4z²

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

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

7x-7y+ax-ay=(7x-7y)+(ax-ay)=7(x-y)+a(x-y)=(x-y)(7+a)

7x-7y+ax-ay

=(7x-7y)+(ax-ay)

=7(x-y)+a(x-y)

=(x-y)(7+a)

Ta có: ( xy+1)^2 - (x+y)^2

= x^2.y^2 + 2xy + 1^2 - x^2 -2xy - y^2

= x^2. y^2 - x^2 - y^2 +1

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

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