a: Ta có: \(m^2+2mn+n^2-p^2+2pq+q^2\)

\(=\left(m+n\right)^2-\left(p-q\right)^2\)

\(=\left(m+n-p+q\right)\left(m+n+p-q\right)\)

a) \(x^2-xy+x-y\)

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

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

b) \(x^2+5x+6\)

\(=x^2+2x+3x+6\)

\(=x\left(x+2\right)+3\left(x+2\right)\)

\(=\left(x+3\right)\left(x+2\right)\)

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

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

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

\(m^2-16+n^2-2mn\)

\(=n^2-2mn+m^2-16\)

\(=\left(n-m\right)^2-16\)

\(=\left(n-m-4\right)\left(n-m+4\right)\)

m² - 16 + n² - 2mn

= m² - 2mn + n² - 16

= (m - n)² - 4²

= (m - n - 4)(m - n + 4)

m² -16 +n² - 2mn = (m² - 2mn + n²) -16

=( m - n )² - 4²

= (m- n- 4)(m - n + 4)

Bài 1: 

\(=3x^3y-6x^2y^2+15xy\)

Bài 2: 

\(=\left(x+y\right)^2-25=\left(x+y+5\right)\left(x+y-5\right)\)

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

\(x^2+2xy+y^2-25y^2\)

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

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

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

\(=\left(x-4y\right).\left(x+6y\right)\)

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

x^2 - 2xy + y^2 - z^2 

=(x-y)²-z²

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

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

a) x³-2x²-x+2

=x(x²-1)+2(-x²+1)

=x(x²-1)-2(x²-1)

=(x²-1)(x-2)

b)

x²+6x-y²+9

=x²+6x+9-y²

=(x+3)²-y²

^{=(x+3-y)(x+3+y)}

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

= - (x² - 2xy + y² - 9) = - [ (x - y)² - 3² ] = - (x - y - 3).(x - y + 3)