ab - ac - bc - c² = -1

a(b - c) - c(b - c) = -1

(a - c)(b -c) = -1

(2- 1)(0 - 1) = -1

< = > a + b = 2 

a) \(\dfrac{x^2-10x+25}{x^2-5x}=0\)

\(\Leftrightarrow x^2-10x+25=0\)

\(\Leftrightarrow x^2-2.x.5+5^2=0\)

\(\Leftrightarrow\left(x-5\right)^2=0\)

\(\Leftrightarrow x-5=0\)

\(\Leftrightarrow x=5\)

Vậy \(x=5\) thì \(\dfrac{x^2-10x+25}{x^2-5x}=0\)

b) \(\dfrac{x^2-10x+25}{x^2-5x}=\dfrac{5}{2}\)

\(\Leftrightarrow\left(x^2-10x+25\right).2=\left(x^2-5x\right).5\)

\(\Leftrightarrow2.\left(x^2-10x+25\right)-5.\left(x^2-5x\right)=0\)

\(\Leftrightarrow\left(2x^2-20x+50\right)-\left(5x^2-25x\right)=0\)

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

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

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

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

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